theory RTree
imports "~~/src/HOL/Library/Transitive_Closure_Table" Max
begin
section {* A theory of relational trees *}
inductive_cases path_nilE [elim!]: "rtrancl_path r x [] y"
inductive_cases path_consE [elim!]: "rtrancl_path r x (z#zs) y"
subsection {* Definitions *}
text {*
In this theory, we are going to give a notion of of `Relational Graph` and
its derived notion `Relational Tree`. Given a binary relation @{text "r"},
the `Relational Graph of @{text "r"}` is the graph, the edges of which
are those in @{text "r"}. In this way, any binary relation can be viewed
as a `Relational Graph`. Note, this notion of graph includes infinite graphs.
A `Relation Graph` @{text "r"} is said to be a `Relational Tree` if it is both
{\em single valued} and {\em acyclic}.
*}
text {*
The following @{text "sgv"} specifies that relation @{text "r"} is {\em single valued}.
*}
locale sgv =
fixes r
assumes sgv: "single_valued r"
text {*
The following @{text "rtree"} specifies that @{text "r"} is a
{\em Relational Tree}.
*}
locale rtree = sgv +
assumes acl: "acyclic r"
text {*
The following two auxiliary functions @{text "rel_of"} and @{text "pred_of"}
transfer between the predicate and set representation of binary relations.
*}
definition "rel_of r = {(x, y) | x y. r x y}"
definition "pred_of r = (\<lambda> x y. (x, y) \<in> r)"
text {*
To reason about {\em Relational Graph}, a notion of path is
needed, which is given by the following @{text "rpath"} (short
for `relational path`).
The path @{text "xs"} in proposition @{text "rpath r x xs y"} is
a path leading from @{text "x"} to @{text "y"}, which serves as a
witness of the fact @{text "(x, y) \<in> r^*"}.
@{text "rpath"}
is simply a wrapper of the @{text "rtrancl_path"} defined in the imported
theory @{text "Transitive_Closure_Table"}, which defines
a notion of path for the predicate form of binary relations.
*}
definition "rpath r x xs y = rtrancl_path (pred_of r) x xs y"
text {*
Given a path @{text "ps"}, @{text "edges_on ps"} is the
set of edges along the path, which is defined as follows:
*}
definition "edges_on ps = {(a,b) | a b. \<exists> xs ys. ps = xs@[a,b]@ys}"
text {*
The following @{text "indep"} defines a notion of independence.
Two nodes @{text "x"} and @{text "y"} are said to be independent
(expressed as @{text "indep x y"}), if neither one is reachable
from the other in relational graph @{text "r"}.
*}
definition "indep r x y = (((x, y) \<notin> r^*) \<and> ((y, x) \<notin> r^*))"
text {*
In relational tree @{text "r"}, the sub tree of node @{text "x"} is written
@{text "subtree r x"}, which is defined to be the set of nodes (including itself)
which can reach @{text "x"} by following some path in @{text "r"}:
*}
definition "subtree r x = {y . (y, x) \<in> r^*}"
definition "ancestors r x = {y. (x, y) \<in> r^+}"
definition "root r x = (ancestors r x = {})"
text {*
The following @{text "edge_in r x"} is the set of edges
contained in the sub-tree of @{text "x"}, with @{text "r"} as the underlying graph.
*}
definition "edges_in r x = {(a, b) | a b. (a, b) \<in> r \<and> b \<in> subtree r x}"
text {*
The following lemma @{text "edges_in_meaning"} shows the intuitive meaning
of `an edge @{text "(a, b)"} is in the sub-tree of @{text "x"}`,
i.e., both @{text "a"} and @{text "b"} are in the sub-tree.
*}
lemma edges_in_meaning:
"edges_in r x = {(a, b) | a b. (a, b) \<in> r \<and> a \<in> subtree r x \<and> b \<in> subtree r x}"
proof -
{ fix a b
assume h: "(a, b) \<in> r" "b \<in> subtree r x"
moreover have "a \<in> subtree r x"
proof -
from h(2)[unfolded subtree_def] have "(b, x) \<in> r^*" by simp
with h(1) have "(a, x) \<in> r^*" by auto
thus ?thesis by (auto simp:subtree_def)
qed
ultimately have "((a, b) \<in> r \<and> a \<in> subtree r x \<and> b \<in> subtree r x)"
by (auto)
} thus ?thesis by (auto simp:edges_in_def)
qed
text {*
The following lemma shows the meaning of @{term "edges_in"} from the other side,
which says: for the edge @{text "(a,b)"} to be outside of the sub-tree of @{text "x"},
it is sufficient to show that @{text "b"} is.
*}
lemma edges_in_refutation:
assumes "b \<notin> subtree r x"
shows "(a, b) \<notin> edges_in r x"
using assms by (unfold edges_in_def subtree_def, auto)
definition "children r x = {y. (y, x) \<in> r}"
locale fbranch =
fixes r
assumes fb: "\<forall> x \<in> Range r . finite (children r x)"
begin
lemma finite_children: "finite (children r x)"
proof(cases "children r x = {}")
case True
thus ?thesis by auto
next
case False
then obtain y where "(y, x) \<in> r" by (auto simp:children_def)
hence "x \<in> Range r" by auto
from fb[rule_format, OF this]
show ?thesis .
qed
end
locale fsubtree = fbranch +
assumes wf: "wf r"
(* ccc *)
subsection {* Auxiliary lemmas *}
lemma index_minimize:
assumes "P (i::nat)"
obtains j where "P j" and "\<forall> k < j. \<not> P k"
proof -
have "\<exists> j. P j \<and> (\<forall> k < j. \<not> P k)"
using assms
proof(induct i rule:less_induct)
case (less t)
show ?case
proof(cases "\<forall> j < t. \<not> P j")
case True
with less (2) show ?thesis by blast
next
case False
then obtain j where "j < t" "P j" by auto
from less(1)[OF this]
show ?thesis .
qed
qed
with that show ?thesis by metis
qed
subsection {* Properties of Relational Graphs and Relational Trees *}
subsubsection {* Properties of @{text "rel_of"} and @{text "pred_of"} *}
text {* The following lemmas establish bijectivity of the two functions *}
lemma pred_rel_eq: "pred_of (rel_of r) = r" by (auto simp:rel_of_def pred_of_def)
lemma rel_pred_eq: "rel_of (pred_of r) = r" by (auto simp:rel_of_def pred_of_def)
lemma rel_of_star: "rel_of (r^**) = (rel_of r)^*"
by (unfold rel_of_def rtranclp_rtrancl_eq, auto)
lemma pred_of_star: "pred_of (r^*) = (pred_of r)^**"
proof -
{ fix x y
have "pred_of (r^*) x y = (pred_of r)^** x y"
by (unfold pred_of_def rtranclp_rtrancl_eq, auto)
} thus ?thesis by auto
qed
lemma star_2_pstar: "(x, y) \<in> r^* = (pred_of (r^*)) x y"
by (simp add: pred_of_def)
subsubsection {* Properties of @{text "rpath"} *}
text {* Induction rule for @{text "rpath"}: *}
lemma rpath_induct [consumes 1, case_names rbase rstep, induct pred: rpath]:
assumes "rpath r x1 x2 x3"
and "\<And>x. P x [] x"
and "\<And>x y ys z. (x, y) \<in> r \<Longrightarrow> rpath r y ys z \<Longrightarrow> P y ys z \<Longrightarrow> P x (y # ys) z"
shows "P x1 x2 x3"
using assms[unfolded rpath_def]
by (induct, auto simp:pred_of_def rpath_def)
lemma rpathE:
assumes "rpath r x xs y"
obtains (base) "y = x" "xs = []"
| (step) z zs where "(x, z) \<in> r" "rpath r z zs y" "xs = z#zs"
using assms
by (induct, auto)
text {* Introduction rule for empty path *}
lemma rbaseI [intro!]:
assumes "x = y"
shows "rpath r x [] y"
by (unfold rpath_def assms,
rule Transitive_Closure_Table.rtrancl_path.base)
text {* Introduction rule for non-empty path *}
lemma rstepI [intro!]:
assumes "(x, y) \<in> r"
and "rpath r y ys z"
shows "rpath r x (y#ys) z"
proof(unfold rpath_def, rule Transitive_Closure_Table.rtrancl_path.step)
from assms(1) show "pred_of r x y" by (auto simp:pred_of_def)
next
from assms(2) show "rtrancl_path (pred_of r) y ys z"
by (auto simp:pred_of_def rpath_def)
qed
text {* Introduction rule for @{text "@"}-path *}
lemma rpath_appendI [intro]:
assumes "rpath r x xs a" and "rpath r a ys y"
shows "rpath r x (xs @ ys) y"
using assms
by (unfold rpath_def, auto intro:rtrancl_path_trans)
text {* Elimination rule for empty path *}
lemma rpath_cases [cases pred:rpath]:
assumes "rpath r a1 a2 a3"
obtains (rbase) "a1 = a3" and "a2 = []"
| (rstep) y :: "'a" and ys :: "'a list"
where "(a1, y) \<in> r" and "a2 = y # ys" and "rpath r y ys a3"
using assms [unfolded rpath_def]
by (cases, auto simp:rpath_def pred_of_def)
lemma rpath_nilE [elim!, cases pred:rpath]:
assumes "rpath r x [] y"
obtains "y = x"
using assms[unfolded rpath_def] by auto
-- {* This is a auxiliary lemmas used only in the proof of @{text "rpath_nnl_lastE"} *}
lemma rpath_nnl_last:
assumes "rtrancl_path r x xs y"
and "xs \<noteq> []"
obtains xs' where "xs = xs'@[y]"
proof -
from append_butlast_last_id[OF `xs \<noteq> []`, symmetric]
obtain xs' y' where eq_xs: "xs = (xs' @ y' # [])" by simp
with assms(1)
have "rtrancl_path r x ... y" by simp
hence "y = y'" by (rule rtrancl_path_appendE, auto)
with eq_xs have "xs = xs'@[y]" by simp
from that[OF this] show ?thesis .
qed
text {*
Elimination rule for non-empty paths constructed with @{text "#"}.
*}
lemma rpath_ConsE [elim!, cases pred:rpath]:
assumes "rpath r x (y # ys) x2"
obtains (rstep) "(x, y) \<in> r" and "rpath r y ys x2"
using assms[unfolded rpath_def]
by (cases, auto simp:rpath_def pred_of_def)
text {*
Elimination rule for non-empty path, where the destination node
@{text "y"} is shown to be at the end of the path.
*}
lemma rpath_nnl_lastE:
assumes "rpath r x xs y"
and "xs \<noteq> []"
obtains xs' where "xs = xs'@[y]"
using assms[unfolded rpath_def]
by (rule rpath_nnl_last, auto)
text {* Other elimination rules of @{text "rpath"} *}
lemma rpath_appendE:
assumes "rpath r x (xs @ [a] @ ys) y"
obtains "rpath r x (xs @ [a]) a" and "rpath r a ys y"
using rtrancl_path_appendE[OF assms[unfolded rpath_def, simplified], folded rpath_def]
by auto
lemma rpath_subE:
assumes "rpath r x (xs @ [a] @ ys @ [b] @ zs) y"
obtains "rpath r x (xs @ [a]) a" and "rpath r a (ys @ [b]) b" and "rpath r b zs y"
using assms
by (elim rpath_appendE, auto)
text {* Every path has a unique end point. *}
lemma rpath_dest_eq:
assumes "rpath r x xs x1"
and "rpath r x xs x2"
shows "x1 = x2"
using assms
by (induct, auto)
subsubsection {* Properites of @{text "edges_on"} *}
lemma edges_on_unfold:
"edges_on (a # b # xs) = {(a, b)} \<union> edges_on (b # xs)" (is "?L = ?R")
proof -
{ fix c d
assume "(c, d) \<in> ?L"
then obtain l1 l2 where h: "(a # b # xs) = l1 @ [c, d] @ l2"
by (auto simp:edges_on_def)
have "(c, d) \<in> ?R"
proof(cases "l1")
case Nil
with h have "(c, d) = (a, b)" by auto
thus ?thesis by auto
next
case (Cons e es)
from h[unfolded this] have "b#xs = es@[c, d]@l2" by auto
thus ?thesis by (auto simp:edges_on_def)
qed
} moreover
{ fix c d
assume "(c, d) \<in> ?R"
moreover have "(a, b) \<in> ?L"
proof -
have "(a # b # xs) = []@[a,b]@xs" by simp
hence "\<exists> l1 l2. (a # b # xs) = l1@[a,b]@l2" by auto
thus ?thesis by (unfold edges_on_def, simp)
qed
moreover {
assume "(c, d) \<in> edges_on (b#xs)"
then obtain l1 l2 where "b#xs = l1@[c, d]@l2" by (unfold edges_on_def, auto)
hence "a#b#xs = (a#l1)@[c,d]@l2" by simp
hence "\<exists> l1 l2. (a # b # xs) = l1@[c,d]@l2" by metis
hence "(c,d) \<in> ?L" by (unfold edges_on_def, simp)
}
ultimately have "(c, d) \<in> ?L" by auto
} ultimately show ?thesis by auto
qed
lemma edges_on_len:
assumes "(a,b) \<in> edges_on l"
shows "length l \<ge> 2"
using assms
by (unfold edges_on_def, auto)
text {* Elimination of @{text "edges_on"} for non-empty path *}
lemma edges_on_consE [elim, cases set:edges_on]:
assumes "(a,b) \<in> edges_on (x#xs)"
obtains (head) xs' where "x = a" and "xs = b#xs'"
| (tail) "(a,b) \<in> edges_on xs"
proof -
from assms obtain l1 l2
where h: "(x#xs) = l1 @ [a,b] @ l2" by (unfold edges_on_def, blast)
have "(\<exists> xs'. x = a \<and> xs = b#xs') \<or> ((a,b) \<in> edges_on xs)"
proof(cases "l1")
case Nil with h
show ?thesis by auto
next
case (Cons e el)
from h[unfolded this]
have "xs = el @ [a,b] @ l2" by auto
thus ?thesis
by (unfold edges_on_def, auto)
qed
thus ?thesis
proof
assume "(\<exists>xs'. x = a \<and> xs = b # xs')"
then obtain xs' where "x = a" "xs = b#xs'" by blast
from that(1)[OF this] show ?thesis .
next
assume "(a, b) \<in> edges_on xs"
from that(2)[OF this] show ?thesis .
qed
qed
text {*
Every edges on the path is a graph edges:
*}
lemma rpath_edges_on:
assumes "rpath r x xs y"
shows "(edges_on (x#xs)) \<subseteq> r"
using assms
proof(induct arbitrary:y)
case (rbase x)
thus ?case by (unfold edges_on_def, auto)
next
case (rstep x y ys z)
show ?case
proof -
{ fix a b
assume "(a, b) \<in> edges_on (x # y # ys)"
hence "(a, b) \<in> r" by (cases, insert rstep, auto)
} thus ?thesis by auto
qed
qed
text {* @{text "edges_on"} is mono with respect to @{text "#"}-operation: *}
lemma edges_on_Cons_mono:
shows "edges_on xs \<subseteq> edges_on (x#xs)"
proof -
{ fix a b
assume "(a, b) \<in> edges_on xs"
then obtain l1 l2 where "xs = l1 @ [a,b] @ l2"
by (auto simp:edges_on_def)
hence "x # xs = (x#l1) @ [a, b] @ l2" by auto
hence "(a, b) \<in> edges_on (x#xs)"
by (unfold edges_on_def, blast)
} thus ?thesis by auto
qed
text {*
The following rule @{text "rpath_transfer"} is used to show
that one path is intact as long as all the edges on it are intact
with the change of graph.
If @{text "x#xs"} is path in graph @{text "r1"} and
every edges along the path is also in @{text "r2"},
then @{text "x#xs"} is also a edge in graph @{text "r2"}:
*}
lemma rpath_transfer:
assumes "rpath r1 x xs y"
and "edges_on (x#xs) \<subseteq> r2"
shows "rpath r2 x xs y"
using assms
proof(induct)
case (rstep x y ys z)
show ?case
proof(rule rstepI)
show "(x, y) \<in> r2"
proof -
have "(x, y) \<in> edges_on (x # y # ys)"
by (unfold edges_on_def, auto)
with rstep(4) show ?thesis by auto
qed
next
show "rpath r2 y ys z"
using rstep edges_on_Cons_mono[of "y#ys" "x"] by (auto)
qed
qed (unfold rpath_def, auto intro!:Transitive_Closure_Table.rtrancl_path.base)
lemma edges_on_rpathI:
assumes "edges_on (a#xs@[b]) \<subseteq> r"
shows "rpath r a (xs@[b]) b"
using assms
proof(induct xs arbitrary: a b)
case Nil
moreover have "(a, b) \<in> edges_on (a # [] @ [b])"
by (unfold edges_on_def, auto)
ultimately have "(a, b) \<in> r" by auto
thus ?case by auto
next
case (Cons x xs a b)
from this(2) have "edges_on (x # xs @ [b]) \<subseteq> r" by (simp add:edges_on_unfold)
from Cons(1)[OF this] have " rpath r x (xs @ [b]) b" .
moreover from Cons(2) have "(a, x) \<in> r" by (auto simp:edges_on_unfold)
ultimately show ?case by (auto)
qed
text {*
The following lemma extracts the path from @{text "x"} to @{text "y"}
from proposition @{text "(x, y) \<in> r^*"}
*}
lemma star_rpath:
assumes "(x, y) \<in> r^*"
obtains xs where "rpath r x xs y"
proof -
have "\<exists> xs. rpath r x xs y"
proof(unfold rpath_def, rule iffD1[OF rtranclp_eq_rtrancl_path])
from assms
show "(pred_of r)\<^sup>*\<^sup>* x y"
apply (fold pred_of_star)
by (auto simp:pred_of_def)
qed
from that and this show ?thesis by blast
qed
text {*
The following lemma uses the path @{text "xs"} from @{text "x"} to @{text "y"}
as a witness to show @{text "(x, y) \<in> r^*"}.
*}
lemma rpath_star:
assumes "rpath r x xs y"
shows "(x, y) \<in> r^*"
proof -
from iffD2[OF rtranclp_eq_rtrancl_path] and assms[unfolded rpath_def]
have "(pred_of r)\<^sup>*\<^sup>* x y" by metis
thus ?thesis by (simp add: pred_of_star star_2_pstar)
qed
lemma subtree_transfer:
assumes "a \<in> subtree r1 a'"
and "r1 \<subseteq> r2"
shows "a \<in> subtree r2 a'"
proof -
from assms(1)[unfolded subtree_def]
have "(a, a') \<in> r1^*" by auto
from star_rpath[OF this]
obtain xs where rp: "rpath r1 a xs a'" by blast
hence "rpath r2 a xs a'"
proof(rule rpath_transfer)
from rpath_edges_on[OF rp] and assms(2)
show "edges_on (a # xs) \<subseteq> r2" by simp
qed
from rpath_star[OF this]
show ?thesis by (auto simp:subtree_def)
qed
lemma subtree_rev_transfer:
assumes "a \<notin> subtree r2 a'"
and "r1 \<subseteq> r2"
shows "a \<notin> subtree r1 a'"
using assms and subtree_transfer by metis
text {*
The following lemmas establishes a relation from paths in @{text "r"}
to @{text "r^+"} relation.
*}
lemma rpath_plus:
assumes "rpath r x xs y"
and "xs \<noteq> []"
shows "(x, y) \<in> r^+"
proof -
from assms(2) obtain e es where "xs = e#es" by (cases xs, auto)
from assms(1)[unfolded this]
show ?thesis
proof(cases)
case rstep
show ?thesis
proof -
from rpath_star[OF rstep(2)] have "(e, y) \<in> r\<^sup>*" .
with rstep(1) show "(x, y) \<in> r^+" by auto
qed
qed
qed
lemma plus_rpath:
assumes "(x, y) \<in> r^+"
obtains xs where "rpath r x xs y" and "xs \<noteq> []"
proof -
from assms
show ?thesis
proof(cases rule:converse_tranclE[consumes 1])
case 1
hence "rpath r x [y] y" by auto
from that[OF this] show ?thesis by auto
next
case (2 z)
from 2(2) have "(z, y) \<in> r^*" by auto
from star_rpath[OF this] obtain xs where "rpath r z xs y" by auto
from rstepI[OF 2(1) this]
have "rpath r x (z # xs) y" .
from that[OF this] show ?thesis by auto
qed
qed
subsubsection {* Properties of @{text "subtree"} and @{term "ancestors"}*}
lemma ancestors_subtreeI:
assumes "b \<in> ancestors r a"
shows "a \<in> subtree r b"
using assms by (auto simp:subtree_def ancestors_def)
lemma ancestors_Field:
assumes "b \<in> ancestors r a"
obtains "a \<in> Domain r" "b \<in> Range r"
using assms
apply (unfold ancestors_def, simp)
by (metis Domain.DomainI Range.intros trancl_domain trancl_range)
lemma subtreeE:
assumes "a \<in> subtree r b"
obtains "a = b"
| "a \<noteq> b" and "b \<in> ancestors r a"
proof -
from assms have "(a, b) \<in> r^*" by (auto simp:subtree_def)
from rtranclD[OF this]
have " a = b \<or> a \<noteq> b \<and> (a, b) \<in> r\<^sup>+" .
with that[unfolded ancestors_def] show ?thesis by auto
qed
lemma subtree_Field:
assumes "a \<in> Field r"
shows "subtree r a \<subseteq> Field r"
by (metis Field_def UnI1 ancestors_Field assms subsetI subtreeE)
lemma subtree_Field:
"subtree r x \<subseteq> Field r \<union> {x}"
proof
fix y
assume "y \<in> subtree r x"
thus "y \<in> Field r \<union> {x}"
proof(cases rule:subtreeE)
case 1
thus ?thesis by auto
next
case 2
thus ?thesis apply (auto simp:ancestors_def)
using Field_def tranclD by fastforce
qed
qed
lemma subtree_ancestorsI:
assumes "a \<in> subtree r b"
and "a \<noteq> b"
shows "b \<in> ancestors r a"
using assms
by (auto elim!:subtreeE)
text {*
@{text "subtree"} is mono with respect to the underlying graph.
*}
lemma subtree_mono:
assumes "r1 \<subseteq> r2"
shows "subtree r1 x \<subseteq> subtree r2 x"
proof
fix c
assume "c \<in> subtree r1 x"
hence "(c, x) \<in> r1^*" by (auto simp:subtree_def)
from star_rpath[OF this] obtain xs
where rp:"rpath r1 c xs x" by metis
hence "rpath r2 c xs x"
proof(rule rpath_transfer)
from rpath_edges_on[OF rp] have "edges_on (c # xs) \<subseteq> r1" .
with assms show "edges_on (c # xs) \<subseteq> r2" by auto
qed
thus "c \<in> subtree r2 x"
by (rule rpath_star[elim_format], auto simp:subtree_def)
qed
text {*
The following lemma characterizes the change of sub-tree of @{text "x"}
with the removal of an outside edge @{text "(a,b)"}.
Note that, according to lemma @{thm edges_in_refutation}, the assumption
@{term "b \<notin> subtree r x"} amounts to saying @{text "(a, b)"}
is outside the sub-tree of @{text "x"}.
*}
lemma subtree_del_outside: (* ddd *)
assumes "b \<notin> subtree r x"
shows "subtree (r - {(a, b)}) x = (subtree r x)"
proof -
{ fix c
assume "c \<in> (subtree r x)"
hence "(c, x) \<in> r^*" by (auto simp:subtree_def)
hence "c \<in> subtree (r - {(a, b)}) x"
proof(rule star_rpath)
fix xs
assume rp: "rpath r c xs x"
show ?thesis
proof -
from rp
have "rpath (r - {(a, b)}) c xs x"
proof(rule rpath_transfer)
from rpath_edges_on[OF rp] have "edges_on (c # xs) \<subseteq> r" .
moreover have "(a, b) \<notin> edges_on (c#xs)"
proof
assume "(a, b) \<in> edges_on (c # xs)"
then obtain l1 l2 where h: "c#xs = l1@[a,b]@l2" by (auto simp:edges_on_def)
hence "tl (c#xs) = tl (l1@[a,b]@l2)" by simp
then obtain l1' where eq_xs_b: "xs = l1'@[b]@l2" by (cases l1, auto)
from rp[unfolded this]
show False
proof(rule rpath_appendE)
assume "rpath r b l2 x"
thus ?thesis
by(rule rpath_star[elim_format], insert assms(1), auto simp:subtree_def)
qed
qed
ultimately show "edges_on (c # xs) \<subseteq> r - {(a,b)}" by auto
qed
thus ?thesis by (rule rpath_star[elim_format], auto simp:subtree_def)
qed
qed
} moreover {
fix c
assume "c \<in> subtree (r - {(a, b)}) x"
moreover have "... \<subseteq> (subtree r x)" by (rule subtree_mono, auto)
ultimately have "c \<in> (subtree r x)" by auto
} ultimately show ?thesis by auto
qed
(* ddd *)
lemma subset_del_subtree_outside: (* ddd *)
assumes "Range r' \<inter> subtree r x = {}"
shows "subtree (r - r') x = (subtree r x)"
proof -
{ fix c
assume "c \<in> (subtree r x)"
hence "(c, x) \<in> r^*" by (auto simp:subtree_def)
hence "c \<in> subtree (r - r') x"
proof(rule star_rpath)
fix xs
assume rp: "rpath r c xs x"
show ?thesis
proof -
from rp
have "rpath (r - r') c xs x"
proof(rule rpath_transfer)
from rpath_edges_on[OF rp] have "edges_on (c # xs) \<subseteq> r" .
moreover {
fix a b
assume h: "(a, b) \<in> r'"
have "(a, b) \<notin> edges_on (c#xs)"
proof
assume "(a, b) \<in> edges_on (c # xs)"
then obtain l1 l2 where "c#xs = (l1@[a])@[b]@l2" by (auto simp:edges_on_def)
hence "tl (c#xs) = tl (l1@[a,b]@l2)" by simp
then obtain l1' where eq_xs_b: "xs = l1'@[b]@l2" by (cases l1, auto)
from rp[unfolded this]
show False
proof(rule rpath_appendE)
assume "rpath r b l2 x"
from rpath_star[OF this]
have "b \<in> subtree r x" by (auto simp:subtree_def)
with assms (1) and h show ?thesis by (auto)
qed
qed
} ultimately show "edges_on (c # xs) \<subseteq> r - r'" by auto
qed
thus ?thesis by (rule rpath_star[elim_format], auto simp:subtree_def)
qed
qed
} moreover {
fix c
assume "c \<in> subtree (r - r') x"
moreover have "... \<subseteq> (subtree r x)" by (rule subtree_mono, auto)
ultimately have "c \<in> (subtree r x)" by auto
} ultimately show ?thesis by auto
qed
lemma subtree_insert_ext:
assumes "b \<in> subtree r x"
shows "subtree (r \<union> {(a, b)}) x = (subtree r x) \<union> (subtree r a)"
using assms by (auto simp:subtree_def rtrancl_insert)
lemma subtree_insert_next:
assumes "b \<notin> subtree r x"
shows "subtree (r \<union> {(a, b)}) x = (subtree r x)"
using assms
by (auto simp:subtree_def rtrancl_insert)
lemma set_add_rootI:
assumes "root r a"
and "a \<notin> Domain r1"
shows "root (r \<union> r1) a"
proof -
let ?r = "r \<union> r1"
{ fix a'
assume "a' \<in> ancestors ?r a"
hence "(a, a') \<in> ?r^+" by (auto simp:ancestors_def)
from tranclD[OF this] obtain z where "(a, z) \<in> ?r" by auto
moreover have "(a, z) \<notin> r"
proof
assume "(a, z) \<in> r"
with assms(1) show False
by (auto simp:root_def ancestors_def)
qed
ultimately have "(a, z) \<in> r1" by auto
with assms(2)
have False by (auto)
} thus ?thesis by (auto simp:root_def)
qed
lemma ancestors_mono:
assumes "r1 \<subseteq> r2"
shows "ancestors r1 x \<subseteq> ancestors r2 x"
proof
fix a
assume "a \<in> ancestors r1 x"
hence "(x, a) \<in> r1^+" by (auto simp:ancestors_def)
from plus_rpath[OF this] obtain xs where
h: "rpath r1 x xs a" "xs \<noteq> []" .
have "rpath r2 x xs a"
proof(rule rpath_transfer[OF h(1)])
from rpath_edges_on[OF h(1)] and assms
show "edges_on (x # xs) \<subseteq> r2" by auto
qed
from rpath_plus[OF this h(2)]
show "a \<in> ancestors r2 x" by (auto simp:ancestors_def)
qed
lemma subtree_refute:
assumes "x \<notin> ancestors r y"
and "x \<noteq> y"
shows "y \<notin> subtree r x"
proof
assume "y \<in> subtree r x"
thus False
by(elim subtreeE, insert assms, auto)
qed
subsubsection {* Properties about relational trees *}
context rtree
begin
lemma ancestors_headE:
assumes "c \<in> ancestors r a"
assumes "(a, b) \<in> r"
obtains "b = c"
| "c \<in> ancestors r b"
proof -
from assms(1)
have "(a, c) \<in> r^+" by (auto simp:ancestors_def)
hence "b = c \<or> c \<in> ancestors r b"
proof(cases rule:converse_tranclE[consumes 1])
case 1
with assms(2) and sgv have "b = c" by (auto simp:single_valued_def)
thus ?thesis by auto
next
case (2 y)
from 2(1) and assms(2) and sgv have "y = b" by (auto simp:single_valued_def)
from 2(2)[unfolded this] have "c \<in> ancestors r b" by (auto simp:ancestors_def)
thus ?thesis by auto
qed
with that show ?thesis by metis
qed
lemma ancestors_accum:
assumes "(a, b) \<in> r"
shows "ancestors r a = ancestors r b \<union> {b}"
proof -
{ fix c
assume "c \<in> ancestors r a"
hence "(a, c) \<in> r^+" by (auto simp:ancestors_def)
hence "c \<in> ancestors r b \<union> {b}"
proof(cases rule:converse_tranclE[consumes 1])
case 1
with sgv assms have "c = b" by (unfold single_valued_def, auto)
thus ?thesis by auto
next
case (2 c')
with sgv assms have "c' = b" by (unfold single_valued_def, auto)
from 2(2)[unfolded this]
show ?thesis by (auto simp:ancestors_def)
qed
} moreover {
fix c
assume "c \<in> ancestors r b \<union> {b}"
hence "c = b \<or> c \<in> ancestors r b" by auto
hence "c \<in> ancestors r a"
proof
assume "c = b"
from assms[folded this]
show ?thesis by (auto simp:ancestors_def)
next
assume "c \<in> ancestors r b"
with assms show ?thesis by (auto simp:ancestors_def)
qed
} ultimately show ?thesis by auto
qed
lemma rootI:
assumes h: "\<And> x'. x' \<noteq> x \<Longrightarrow> x \<notin> subtree r' x'"
and "r' \<subseteq> r"
shows "root r' x"
proof -
from acyclic_subset[OF acl assms(2)]
have acl': "acyclic r'" .
{ fix x'
assume "x' \<in> ancestors r' x"
hence h1: "(x, x') \<in> r'^+" by (auto simp:ancestors_def)
have "x' \<noteq> x"
proof
assume eq_x: "x' = x"
from h1[unfolded this] and acl'
show False by (auto simp:acyclic_def)
qed
moreover from h1 have "x \<in> subtree r' x'" by (auto simp:subtree_def)
ultimately have False using h by auto
} thus ?thesis by (auto simp:root_def)
qed
lemma rpath_overlap_oneside: (* ddd *)
assumes "rpath r x xs1 x1"
and "rpath r x xs2 x2"
and "length xs1 \<le> length xs2"
obtains xs3 where "xs2 = xs1 @ xs3"
proof(cases "xs1 = []")
case True
with that show ?thesis by auto
next
case False
have "\<forall> i \<le> length xs1. take i xs1 = take i xs2"
proof -
{ assume "\<not> (\<forall> i \<le> length xs1. take i xs1 = take i xs2)"
then obtain i where "i \<le> length xs1 \<and> take i xs1 \<noteq> take i xs2" by auto
from this(1) have "False"
proof(rule index_minimize)
fix j
assume h1: "j \<le> length xs1 \<and> take j xs1 \<noteq> take j xs2"
and h2: " \<forall>k<j. \<not> (k \<le> length xs1 \<and> take k xs1 \<noteq> take k xs2)"
-- {* @{text "j - 1"} is the branch point between @{text "xs1"} and @{text "xs2"} *}
let ?idx = "j - 1"
-- {* A number of inequalities concerning @{text "j - 1"} are derived first *}
have lt_i: "?idx < length xs1" using False h1
by (metis Suc_diff_1 le_neq_implies_less length_greater_0_conv lessI less_imp_diff_less)
have lt_i': "?idx < length xs2" using lt_i and assms(3) by auto
have lt_j: "?idx < j" using h1 by (cases j, auto)
-- {* From thesis inequalities, a number of equations concerning @{text "xs1"}
and @{text "xs2"} are derived *}
have eq_take: "take ?idx xs1 = take ?idx xs2"
using h2[rule_format, OF lt_j] and h1 by auto
have eq_xs1: " xs1 = take ?idx xs1 @ xs1 ! (?idx) # drop (Suc (?idx)) xs1"
using id_take_nth_drop[OF lt_i] .
have eq_xs2: "xs2 = take ?idx xs2 @ xs2 ! (?idx) # drop (Suc (?idx)) xs2"
using id_take_nth_drop[OF lt_i'] .
-- {* The branch point along the path is finally pinpointed *}
have neq_idx: "xs1!?idx \<noteq> xs2!?idx"
proof -
have "take j xs1 = take ?idx xs1 @ [xs1 ! ?idx]"
using eq_xs1 Suc_diff_1 lt_i lt_j take_Suc_conv_app_nth by fastforce
moreover have eq_tk2: "take j xs2 = take ?idx xs2 @ [xs2 ! ?idx]"
using Suc_diff_1 lt_i' lt_j take_Suc_conv_app_nth by fastforce
ultimately show ?thesis using eq_take h1 by auto
qed
show ?thesis
proof(cases " take (j - 1) xs1 = []")
case True
have "(x, xs1!?idx) \<in> r"
proof -
from eq_xs1[unfolded True, simplified, symmetric] assms(1)
have "rpath r x ( xs1 ! ?idx # drop (Suc ?idx) xs1) x1" by simp
from this[unfolded rpath_def]
show ?thesis by (auto simp:pred_of_def)
qed
moreover have "(x, xs2!?idx) \<in> r"
proof -
from eq_xs2[folded eq_take, unfolded True, simplified, symmetric] assms(2)
have "rpath r x ( xs2 ! ?idx # drop (Suc ?idx) xs2) x2" by simp
from this[unfolded rpath_def]
show ?thesis by (auto simp:pred_of_def)
qed
ultimately show ?thesis using neq_idx sgv[unfolded single_valued_def] by metis
next
case False
then obtain e es where eq_es: "take ?idx xs1 = es@[e]"
using rev_exhaust by blast
have "(e, xs1!?idx) \<in> r"
proof -
from eq_xs1[unfolded eq_es]
have "xs1 = es@[e, xs1!?idx]@drop (Suc ?idx) xs1" by simp
hence "(e, xs1!?idx) \<in> edges_on xs1" by (simp add:edges_on_def, metis)
with rpath_edges_on[OF assms(1)] edges_on_Cons_mono[of xs1 x]
show ?thesis by auto
qed moreover have "(e, xs2!?idx) \<in> r"
proof -
from eq_xs2[folded eq_take, unfolded eq_es]
have "xs2 = es@[e, xs2!?idx]@drop (Suc ?idx) xs2" by simp
hence "(e, xs2!?idx) \<in> edges_on xs2" by (simp add:edges_on_def, metis)
with rpath_edges_on[OF assms(2)] edges_on_Cons_mono[of xs2 x]
show ?thesis by auto
qed
ultimately show ?thesis
using sgv[unfolded single_valued_def] neq_idx by metis
qed
qed
} thus ?thesis by auto
qed
from this[rule_format, of "length xs1"]
have "take (length xs1) xs1 = take (length xs1) xs2" by simp
moreover have "xs2 = take (length xs1) xs2 @ drop (length xs1) xs2" by simp
ultimately have "xs2 = xs1 @ drop (length xs1) xs2" by auto
from that[OF this] show ?thesis .
qed
lemma rpath_overlap [consumes 2, cases pred:rpath]:
assumes "rpath r x xs1 x1"
and "rpath r x xs2 x2"
obtains (less_1) xs3 where "xs2 = xs1 @ xs3"
| (less_2) xs3 where "xs1 = xs2 @ xs3"
proof -
have "length xs1 \<le> length xs2 \<or> length xs2 \<le> length xs1" by auto
with assms rpath_overlap_oneside that show ?thesis by metis
qed
text {*
As a corollary of @{thm "rpath_overlap_oneside"},
the following two lemmas gives one important property of relation tree,
i.e. there is at most one path between any two nodes.
Similar to the proof of @{thm rpath_overlap}, we starts with
the one side version first.
*}
lemma rpath_unique_oneside:
assumes "rpath r x xs1 y"
and "rpath r x xs2 y"
and "length xs1 \<le> length xs2"
shows "xs1 = xs2"
proof -
from rpath_overlap_oneside[OF assms]
obtain xs3 where less_1: "xs2 = xs1 @ xs3" by blast
show ?thesis
proof(cases "xs3 = []")
case True
from less_1[unfolded this] show ?thesis by simp
next
case False
note FalseH = this
show ?thesis
proof(cases "xs1 = []")
case True
have "(x, x) \<in> r^+"
proof(rule rpath_plus)
from assms(1)[unfolded True]
have "y = x" by (cases rule:rpath_nilE, simp)
from assms(2)[unfolded this] show "rpath r x xs2 x" .
next
from less_1 and False show "xs2 \<noteq> []" by simp
qed
with acl show ?thesis by (unfold acyclic_def, auto)
next
case False
then obtain e es where eq_xs1: "xs1 = es@[e]" using rev_exhaust by auto
from assms(2)[unfolded less_1 this]
have "rpath r x (es @ [e] @ xs3) y" by simp
thus ?thesis
proof(cases rule:rpath_appendE)
case 1
from rpath_dest_eq [OF 1(1)[folded eq_xs1] assms(1)]
have "e = y" .
from rpath_plus [OF 1(2)[unfolded this] FalseH]
have "(y, y) \<in> r^+" .
with acl show ?thesis by (unfold acyclic_def, auto)
qed
qed
qed
qed
text {*
The following is the full version of path uniqueness.
*}
lemma rpath_unique:
assumes "rpath r x xs1 y"
and "rpath r x xs2 y"
shows "xs1 = xs2"
proof(cases "length xs1 \<le> length xs2")
case True
from rpath_unique_oneside[OF assms this] show ?thesis .
next
case False
hence "length xs2 \<le> length xs1" by simp
from rpath_unique_oneside[OF assms(2,1) this]
show ?thesis by simp
qed
text {*
The following lemma shows that the `independence` relation is symmetric.
It is an obvious auxiliary lemma which will be used later.
*}
lemma sym_indep: "indep r x y \<Longrightarrow> indep r y x"
by (unfold indep_def, auto)
text {*
This is another `obvious` lemma about trees, which says trees rooted at
independent nodes are disjoint.
*}
lemma subtree_disjoint:
assumes "indep r x y"
shows "subtree r x \<inter> subtree r y = {}"
proof -
{ fix z x y xs1 xs2 xs3
assume ind: "indep r x y"
and rp1: "rpath r z xs1 x"
and rp2: "rpath r z xs2 y"
and h: "xs2 = xs1 @ xs3"
have False
proof(cases "xs1 = []")
case True
from rp1[unfolded this] have "x = z" by auto
from rp2[folded this] rpath_star ind[unfolded indep_def]
show ?thesis by metis
next
case False
then obtain e es where eq_xs1: "xs1 = es@[e]" using rev_exhaust by blast
from rp2[unfolded h this]
have "rpath r z (es @ [e] @ xs3) y" by simp
thus ?thesis
proof(cases rule:rpath_appendE)
case 1
have "e = x" using 1(1)[folded eq_xs1] rp1 rpath_dest_eq by metis
from rpath_star[OF 1(2)[unfolded this]] ind[unfolded indep_def]
show ?thesis by auto
qed
qed
} note my_rule = this
{ fix z
assume h: "z \<in> subtree r x" "z \<in> subtree r y"
from h(1) have "(z, x) \<in> r^*" by (unfold subtree_def, auto)
then obtain xs1 where rp1: "rpath r z xs1 x" using star_rpath by metis
from h(2) have "(z, y) \<in> r^*" by (unfold subtree_def, auto)
then obtain xs2 where rp2: "rpath r z xs2 y" using star_rpath by metis
from rp1 rp2
have False
by (cases, insert my_rule[OF sym_indep[OF assms(1)] rp2 rp1]
my_rule[OF assms(1) rp1 rp2], auto)
} thus ?thesis by auto
qed
text {*
The following lemma @{text "subtree_del"} characterizes the change of sub-tree of
@{text "x"} with the removal of an inside edge @{text "(a, b)"}.
Note that, the case for the removal of an outside edge has already been dealt with
in lemma @{text "subtree_del_outside"}).
This lemma is underpinned by the following two `obvious` facts:
\begin{enumearte}
\item
In graph @{text "r"}, for an inside edge @{text "(a,b) \<in> edges_in r x"},
every node @{text "c"} in the sub-tree of @{text "a"} has a path
which goes first from @{text "c"} to @{text "a"}, then through edge @{text "(a, b)"}, and
finally reaches @{text "x"}. By the uniqueness of path in a tree,
all paths from sub-tree of @{text "a"} to @{text "x"} are such constructed, therefore
must go through @{text "(a, b)"}. The consequence is: with the removal of @{text "(a,b)"},
all such paths will be broken.
\item
On the other hand, all paths not originate from within the sub-tree of @{text "a"}
will not be affected by the removal of edge @{text "(a, b)"}.
The reason is simple: if the path is affected by the removal, it must
contain @{text "(a, b)"}, then it must originate from within the sub-tree of @{text "a"}.
\end{enumearte}
*}
lemma subtree_del_inside: (* ddd *)
assumes "(a,b) \<in> edges_in r x"
shows "subtree (r - {(a, b)}) x = (subtree r x) - subtree r a"
proof -
from assms have asm: "b \<in> subtree r x" "(a, b) \<in> r" by (auto simp:edges_in_def)
-- {* The proof follows a common pattern to prove the equality of sets. *}
{ -- {* The `left to right` direction.
*}
fix c
-- {* Assuming @{text "c"} is inside the sub-tree of @{text "x"} in the reduced graph *}
assume h: "c \<in> subtree (r - {(a, b)}) x"
-- {* We are going to show that @{text "c"} can not be in the sub-tree of @{text "a"} in
the original graph. *}
-- {* In other words, all nodes inside the sub-tree of @{text "a"} in the original
graph will be removed from the sub-tree of @{text "x"} in the reduced graph. *}
-- {* The reason, as analyzed before, is that all paths from within the
sub-tree of @{text "a"} are broken with the removal of edge @{text "(a,b)"}.
*}
have "c \<in> (subtree r x) - subtree r a"
proof -
let ?r' = "r - {(a, b)}" -- {* The reduced graph is abbreviated as @{text "?r'"} *}
from h have "(c, x) \<in> ?r'^*" by (auto simp:subtree_def)
-- {* Extract from the reduced graph the path @{text "xs"} from @{text "c"} to @{text "x"}. *}
then obtain xs where rp0: "rpath ?r' c xs x" by (rule star_rpath, auto)
-- {* It is easy to show @{text "xs"} is also a path in the original graph *}
hence rp1: "rpath r c xs x"
proof(rule rpath_transfer)
from rpath_edges_on[OF rp0]
show "edges_on (c # xs) \<subseteq> r" by auto
qed
-- {* @{text "xs"} is used as the witness to show that @{text "c"}
in the sub-tree of @{text "x"} in the original graph. *}
hence "c \<in> subtree r x"
by (rule rpath_star[elim_format], auto simp:subtree_def)
-- {* The next step is to show that @{text "c"} can not be in the sub-tree of @{text "a"}
in the original graph. *}
-- {* We need to use the fact that all paths originate from within sub-tree of @{text "a"}
are broken. *}
moreover have "c \<notin> subtree r a"
proof
-- {* Proof by contradiction, suppose otherwise *}
assume otherwise: "c \<in> subtree r a"
-- {* Then there is a path in original graph leading from @{text "c"} to @{text "a"} *}
obtain xs1 where rp_c: "rpath r c xs1 a"
proof -
from otherwise have "(c, a) \<in> r^*" by (auto simp:subtree_def)
thus ?thesis by (rule star_rpath, auto intro!:that)
qed
-- {* Starting from this path, we are going to construct a fictional
path from @{text "c"} to @{text "x"}, which, as explained before,
is broken, so that contradiction can be derived. *}
-- {* First, there is a path from @{text "b"} to @{text "x"} *}
obtain ys where rp_b: "rpath r b ys x"
proof -
from asm have "(b, x) \<in> r^*" by (auto simp:subtree_def)
thus ?thesis by (rule star_rpath, auto intro!:that)
qed
-- {* The paths @{text "xs1"} and @{text "ys"} can be
tied together using @{text "(a,b)"} to form a path
from @{text "c"} to @{text "x"}: *}
have "rpath r c (xs1 @ b # ys) x"
proof -
from rstepI[OF asm(2) rp_b] have "rpath r a (b # ys) x" .
from rpath_appendI[OF rp_c this]
show ?thesis .
qed
-- {* By the uniqueness of path between two nodes of a tree, we have: *}
from rpath_unique[OF rp1 this] have eq_xs: "xs = xs1 @ b # ys" .
-- {* Contradiction can be derived from from this fictional path . *}
show False
proof -
-- {* It can be shown that @{term "(a,b)"} is on this fictional path. *}
have "(a, b) \<in> edges_on (c#xs)"
proof(cases "xs1 = []")
case True
from rp_c[unfolded this] have "rpath r c [] a" .
hence eq_c: "c = a" by (rule rpath_nilE, simp)
hence "c#xs = a#xs" by simp
from this and eq_xs have "c#xs = a # xs1 @ b # ys" by simp
from this[unfolded True] have "c#xs = []@[a,b]@ys" by simp
thus ?thesis by (auto simp:edges_on_def)
next
case False
from rpath_nnl_lastE[OF rp_c this]
obtain xs' where "xs1 = xs'@[a]" by auto
from eq_xs[unfolded this] have "c#xs = (c#xs')@[a,b]@ys" by simp
thus ?thesis by (unfold edges_on_def, blast)
qed
-- {* It can also be shown that @{term "(a,b)"} is not on this fictional path. *}
moreover have "(a, b) \<notin> edges_on (c#xs)"
using rpath_edges_on[OF rp0] by auto
-- {* Contradiction is thus derived. *}
ultimately show False by auto
qed
qed
ultimately show ?thesis by auto
qed
} moreover {
-- {* The `right to left` direction.
*}
fix c
-- {* Assuming that @{text "c"} is in the sub-tree of @{text "x"}, but
outside of the sub-tree of @{text "a"} in the original graph, *}
assume h: "c \<in> (subtree r x) - subtree r a"
-- {* we need to show that in the reduced graph, @{text "c"} is still in
the sub-tree of @{text "x"}. *}
have "c \<in> subtree (r - {(a, b)}) x"
proof -
-- {* The proof goes by showing that the path from @{text "c"} to @{text "x"}
in the original graph is not affected by the removal of @{text "(a,b)"}.
*}
from h have "(c, x) \<in> r^*" by (unfold subtree_def, auto)
-- {* Extract the path @{text "xs"} from @{text "c"} to @{text "x"} in the original graph. *}
from star_rpath[OF this] obtain xs where rp: "rpath r c xs x" by auto
-- {* Show that it is also a path in the reduced graph. *}
hence "rpath (r - {(a, b)}) c xs x"
-- {* The proof goes by using rule @{thm rpath_transfer} *}
proof(rule rpath_transfer)
-- {* We need to show all edges on the path are still in the reduced graph. *}
show "edges_on (c # xs) \<subseteq> r - {(a, b)}"
proof -
-- {* It is easy to show that all the edges are in the original graph. *}
from rpath_edges_on [OF rp] have " edges_on (c # xs) \<subseteq> r" .
-- {* The essential part is to show that @{text "(a, b)"} is not on the path. *}
moreover have "(a,b) \<notin> edges_on (c#xs)"
proof
-- {* Proof by contradiction, suppose otherwise: *}
assume otherwise: "(a, b) \<in> edges_on (c#xs)"
-- {* Then @{text "(a, b)"} is in the middle of the path.
with @{text "l1"} and @{text "l2"} be the nodes in
the front and rear respectively. *}
then obtain l1 l2 where eq_xs:
"c#xs = l1 @ [a, b] @ l2" by (unfold edges_on_def, blast)
-- {* From this, it can be shown that @{text "c"} is
in the sub-tree of @{text "a"} *}
have "c \<in> subtree r a"
proof(cases "l1 = []")
case True
-- {* If @{text "l1"} is null, it can be derived that @{text "c = a"}. *}
with eq_xs have "c = a" by auto
-- {* So, @{text "c"} is obviously in the sub-tree of @{text "a"}. *}
thus ?thesis by (unfold subtree_def, auto)
next
case False
-- {* When @{text "l1"} is not null, it must have a tail @{text "es"}: *}
then obtain e es where "l1 = e#es" by (cases l1, auto)
-- {* The relation of this tail with @{text "xs"} is derived: *}
with eq_xs have "xs = es@[a,b]@l2" by auto
-- {* From this, a path from @{text "c"} to @{text "a"} is made visible: *}
from rp[unfolded this] have "rpath r c (es @ [a] @ (b#l2)) x" by simp
thus ?thesis
proof(cases rule:rpath_appendE)
-- {* The path from @{text "c"} to @{text "a"} is extraced
using @{thm "rpath_appendE"}: *}
case 1
from rpath_star[OF this(1)]
-- {* The extracted path servers as a witness that @{text "c"} is
in the sub-tree of @{text "a"}: *}
show ?thesis by (simp add:subtree_def)
qed
qed with h show False by auto
qed ultimately show ?thesis by auto
qed
qed
-- {* From , it is shown that @{text "c"} is in the sub-tree of @{text "x"}
inthe reduced graph. *}
from rpath_star[OF this] show ?thesis by (auto simp:subtree_def)
qed
}
-- {* The equality of sets is derived from the two directions just proved. *}
ultimately show ?thesis by auto
qed
lemma set_del_rootI:
assumes "r1 \<subseteq> r"
and "a \<in> Domain r1"
shows "root (r - r1) a"
proof -
let ?r = "r - r1"
{ fix a'
assume neq: "a' \<noteq> a"
have "a \<notin> subtree ?r a'"
proof
assume "a \<in> subtree ?r a'"
hence "(a, a') \<in> ?r^*" by (auto simp:subtree_def)
from star_rpath[OF this] obtain xs
where rp: "rpath ?r a xs a'" by auto
from rpathE[OF this] and neq
obtain z zs where h: "(a, z) \<in> ?r" "rpath ?r z zs a'" "xs = z#zs" by auto
from assms(2) obtain z' where z'_in: "(a, z') \<in> r1" by (auto simp:DomainE)
with assms(1) have "(a, z') \<in> r" by auto
moreover from h(1) have "(a, z) \<in> r" by simp
ultimately have "z' = z" using sgv by (auto simp:single_valued_def)
from z'_in[unfolded this] and h(1) show False by auto
qed
} thus ?thesis by (intro rootI, auto)
qed
lemma edge_del_no_rootI:
assumes "(a, b) \<in> r"
shows "root (r - {(a, b)}) a"
by (rule set_del_rootI, insert assms, auto)
lemma ancestors_children_unique:
assumes "z1 \<in> ancestors r x \<inter> children r y"
and "z2 \<in> ancestors r x \<inter> children r y"
shows "z1 = z2"
proof -
from assms have h:
"(x, z1) \<in> r^+" "(z1, y) \<in> r"
"(x, z2) \<in> r^+" "(z2, y) \<in> r"
by (auto simp:ancestors_def children_def)
-- {* From this, a path containing @{text "z1"} is obtained. *}
from plus_rpath[OF h(1)] obtain xs1
where h1: "rpath r x xs1 z1" "xs1 \<noteq> []" by auto
from rpath_nnl_lastE[OF this] obtain xs1' where eq_xs1: "xs1 = xs1' @ [z1]"
by auto
from h(2) have h2: "rpath r z1 [y] y" by auto
from rpath_appendI[OF h1(1) h2, unfolded eq_xs1]
have rp1: "rpath r x (xs1' @ [z1, y]) y" by simp
-- {* Then, another path containing @{text "z2"} is obtained. *}
from plus_rpath[OF h(3)] obtain xs2
where h3: "rpath r x xs2 z2" "xs2 \<noteq> []" by auto
from rpath_nnl_lastE[OF this] obtain xs2' where eq_xs2: "xs2 = xs2' @ [z2]"
by auto
from h(4) have h4: "rpath r z2 [y] y" by auto
from rpath_appendI[OF h3(1) h4, unfolded eq_xs2]
have "rpath r x (xs2' @ [z2, y]) y" by simp
-- {* Finally @{text "z1 = z2"} is proved by uniqueness of path. *}
from rpath_unique[OF rp1 this]
have "xs1' @ [z1, y] = xs2' @ [z2, y]" .
thus ?thesis by auto
qed
lemma ancestors_childrenE:
assumes "y \<in> ancestors r x"
obtains "x \<in> children r y"
| z where "z \<in> ancestors r x \<inter> children r y"
proof -
from assms(1) have "(x, y) \<in> r^+" by (auto simp:ancestors_def)
from tranclD2[OF this] obtain z where
h: "(x, z) \<in> r\<^sup>*" "(z, y) \<in> r" by auto
from h(1)
show ?thesis
proof(cases rule:rtranclE)
case base
from h(2)[folded this] have "x \<in> children r y"
by (auto simp:children_def)
thus ?thesis by (intro that, auto)
next
case (step u)
hence "z \<in> ancestors r x" by (auto simp:ancestors_def)
moreover from h(2) have "z \<in> children r y"
by (auto simp:children_def)
ultimately show ?thesis by (intro that, auto)
qed
qed
end (* of rtree *)
lemma subtree_children:
"subtree r x = {x} \<union> (\<Union> (subtree r ` (children r x)))" (is "?L = ?R")
proof -
{ fix z
assume "z \<in> ?L"
hence "z \<in> ?R"
proof(cases rule:subtreeE[consumes 1])
case 2
hence "(z, x) \<in> r^+" by (auto simp:ancestors_def)
thus ?thesis
proof(rule tranclE)
assume "(z, x) \<in> r"
hence "z \<in> children r x" by (unfold children_def, auto)
moreover have "z \<in> subtree r z" by (auto simp:subtree_def)
ultimately show ?thesis by auto
next
fix c
assume h: "(z, c) \<in> r\<^sup>+" "(c, x) \<in> r"
hence "c \<in> children r x" by (auto simp:children_def)
moreover from h have "z \<in> subtree r c" by (auto simp:subtree_def)
ultimately show ?thesis by auto
qed
qed auto
} moreover {
fix z
assume h: "z \<in> ?R"
have "x \<in> subtree r x" by (auto simp:subtree_def)
moreover {
assume "z \<in> \<Union>(subtree r ` children r x)"
then obtain y where "(y, x) \<in> r" "(z, y) \<in> r^*"
by (auto simp:subtree_def children_def)
hence "(z, x) \<in> r^*" by auto
hence "z \<in> ?L" by (auto simp:subtree_def)
} ultimately have "z \<in> ?L" using h by auto
} ultimately show ?thesis by auto
qed
context fsubtree
begin
lemma finite_subtree:
shows "finite (subtree r x)"
proof(induct rule:wf_induct[OF wf])
case (1 x)
have "finite (\<Union>(subtree r ` children r x))"
proof(rule finite_Union)
show "finite (subtree r ` children r x)"
proof(cases "children r x = {}")
case True
thus ?thesis by auto
next
case False
hence "x \<in> Range r" by (auto simp:children_def)
from fb[rule_format, OF this]
have "finite (children r x)" .
thus ?thesis by (rule finite_imageI)
qed
next
fix M
assume "M \<in> subtree r ` children r x"
then obtain y where h: "y \<in> children r x" "M = subtree r y" by auto
hence "(y, x) \<in> r" by (auto simp:children_def)
from 1[rule_format, OF this, folded h(2)]
show "finite M" .
qed
thus ?case
by (unfold subtree_children finite_Un, auto)
qed
end
definition "pairself f = (\<lambda>(a, b). (f a, f b))"
definition "rel_map f r = (pairself f ` r)"
lemma rel_mapE:
assumes "(a, b) \<in> rel_map f r"
obtains c d
where "(c, d) \<in> r" "(a, b) = (f c, f d)"
using assms
by (unfold rel_map_def pairself_def, auto)
lemma rel_mapI:
assumes "(a, b) \<in> r"
and "c = f a"
and "d = f b"
shows "(c, d) \<in> rel_map f r"
using assms
by (unfold rel_map_def pairself_def, auto)
lemma map_appendE:
assumes "map f zs = xs @ ys"
obtains xs' ys'
where "zs = xs' @ ys'" "xs = map f xs'" "ys = map f ys'"
proof -
have "\<exists> xs' ys'. zs = xs' @ ys' \<and> xs = map f xs' \<and> ys = map f ys'"
using assms
proof(induct xs arbitrary:zs ys)
case (Nil zs ys)
thus ?case by auto
next
case (Cons x xs zs ys)
note h = this
show ?case
proof(cases zs)
case (Cons e es)
with h have eq_x: "map f es = xs @ ys" "x = f e" by auto
from h(1)[OF this(1)]
obtain xs' ys' where "es = xs' @ ys'" "xs = map f xs'" "ys = map f ys'"
by blast
with Cons eq_x
have "zs = (e#xs') @ ys' \<and> x # xs = map f (e#xs') \<and> ys = map f ys'" by auto
thus ?thesis by metis
qed (insert h, auto)
qed
thus ?thesis by (auto intro!:that)
qed
lemma rel_map_mono:
assumes "r1 \<subseteq> r2"
shows "rel_map f r1 \<subseteq> rel_map f r2"
using assms
by (auto simp:rel_map_def pairself_def)
lemma rel_map_compose [simp]:
shows "rel_map f1 (rel_map f2 r) = rel_map (f1 o f2) r"
by (auto simp:rel_map_def pairself_def)
lemma edges_on_map: "edges_on (map f xs) = rel_map f (edges_on xs)"
proof -
{ fix a b
assume "(a, b) \<in> edges_on (map f xs)"
then obtain l1 l2 where eq_map: "map f xs = l1 @ [a, b] @ l2"
by (unfold edges_on_def, auto)
hence "(a, b) \<in> rel_map f (edges_on xs)"
by (auto elim!:map_appendE intro!:rel_mapI simp:edges_on_def)
} moreover {
fix a b
assume "(a, b) \<in> rel_map f (edges_on xs)"
then obtain c d where
h: "(c, d) \<in> edges_on xs" "(a, b) = (f c, f d)"
by (elim rel_mapE, auto)
then obtain l1 l2 where
eq_xs: "xs = l1 @ [c, d] @ l2"
by (auto simp:edges_on_def)
hence eq_map: "map f xs = map f l1 @ [f c, f d] @ map f l2" by auto
have "(a, b) \<in> edges_on (map f xs)"
proof -
from h(2) have "[f c, f d] = [a, b]" by simp
from eq_map[unfolded this] show ?thesis by (auto simp:edges_on_def)
qed
} ultimately show ?thesis by auto
qed
lemma image_id:
assumes "\<And> x. x \<in> A \<Longrightarrow> f x = x"
shows "f ` A = A"
using assms by (auto simp:image_def)
lemma rel_map_inv_id:
assumes "inj_on f ((Domain r) \<union> (Range r))"
shows "(rel_map (inv_into ((Domain r) \<union> (Range r)) f \<circ> f) r) = r"
proof -
let ?f = "(inv_into (Domain r \<union> Range r) f \<circ> f)"
{
fix a b
assume h0: "(a, b) \<in> r"
have "pairself ?f (a, b) = (a, b)"
proof -
from assms h0 have "?f a = a" by (auto intro:inv_into_f_f)
moreover have "?f b = b"
by (insert h0, simp, intro inv_into_f_f[OF assms], auto intro!:RangeI)
ultimately show ?thesis by (auto simp:pairself_def)
qed
} thus ?thesis by (unfold rel_map_def, intro image_id, case_tac x, auto)
qed
lemma rel_map_acyclic:
assumes "acyclic r"
and "inj_on f ((Domain r) \<union> (Range r))"
shows "acyclic (rel_map f r)"
proof -
let ?D = "Domain r \<union> Range r"
{ fix a
assume "(a, a) \<in> (rel_map f r)^+"
from plus_rpath[OF this]
obtain xs where rp: "rpath (rel_map f r) a xs a" "xs \<noteq> []" by auto
from rpath_nnl_lastE[OF this] obtain xs' where eq_xs: "xs = xs'@[a]" by auto
from rpath_edges_on[OF rp(1)]
have h: "edges_on (a # xs) \<subseteq> rel_map f r" .
from edges_on_map[of "inv_into ?D f" "a#xs"]
have "edges_on (map (inv_into ?D f) (a # xs)) = rel_map (inv_into ?D f) (edges_on (a # xs))" .
with rel_map_mono[OF h, of "inv_into ?D f"]
have "edges_on (map (inv_into ?D f) (a # xs)) \<subseteq> rel_map ((inv_into ?D f) o f) r" by simp
from this[unfolded eq_xs]
have subr: "edges_on (map (inv_into ?D f) (a # xs' @ [a])) \<subseteq> rel_map (inv_into ?D f \<circ> f) r" .
have "(map (inv_into ?D f) (a # xs' @ [a])) = (inv_into ?D f a) # map (inv_into ?D f) xs' @ [inv_into ?D f a]"
by simp
from edges_on_rpathI[OF subr[unfolded this]]
have "rpath (rel_map (inv_into ?D f \<circ> f) r)
(inv_into ?D f a) (map (inv_into ?D f) xs' @ [inv_into ?D f a]) (inv_into ?D f a)" .
hence "(inv_into ?D f a, inv_into ?D f a) \<in> (rel_map (inv_into ?D f \<circ> f) r)^+"
by (rule rpath_plus, simp)
moreover have "(rel_map (inv_into ?D f \<circ> f) r) = r" by (rule rel_map_inv_id[OF assms(2)])
moreover note assms(1)
ultimately have False by (unfold acyclic_def, auto)
} thus ?thesis by (auto simp:acyclic_def)
qed
lemma relpow_mult:
"((r::'a rel) ^^ m) ^^ n = r ^^ (m*n)"
proof(induct n arbitrary:m)
case (Suc k m)
thus ?case
proof -
have h: "(m * k + m) = (m + m * k)" by auto
show ?thesis
apply (simp add:Suc relpow_add[symmetric])
by (unfold h, simp)
qed
qed simp
lemma compose_relpow_2:
assumes "r1 \<subseteq> r"
and "r2 \<subseteq> r"
shows "r1 O r2 \<subseteq> r ^^ (2::nat)"
proof -
{ fix a b
assume "(a, b) \<in> r1 O r2"
then obtain e where "(a, e) \<in> r1" "(e, b) \<in> r2"
by auto
with assms have "(a, e) \<in> r" "(e, b) \<in> r" by auto
hence "(a, b) \<in> r ^^ (Suc (Suc 0))" by auto
} thus ?thesis by (auto simp:numeral_2_eq_2)
qed
lemma acyclic_compose:
assumes "acyclic r"
and "r1 \<subseteq> r"
and "r2 \<subseteq> r"
shows "acyclic (r1 O r2)"
proof -
{ fix a
assume "(a, a) \<in> (r1 O r2)^+"
from trancl_mono[OF this compose_relpow_2[OF assms(2, 3)]]
have "(a, a) \<in> (r ^^ 2) ^+" .
from trancl_power[THEN iffD1, OF this]
obtain n where h: "(a, a) \<in> (r ^^ 2) ^^ n" "n > 0" by blast
from this(1)[unfolded relpow_mult] have h2: "(a, a) \<in> r ^^ (2 * n)" .
have "(a, a) \<in> r^+"
proof(cases rule:trancl_power[THEN iffD2])
from h(2) h2 show "\<exists>n>0. (a, a) \<in> r ^^ n"
by (rule_tac x = "2*n" in exI, auto)
qed
with assms have "False" by (auto simp:acyclic_def)
} thus ?thesis by (auto simp:acyclic_def)
qed
lemma children_compose_unfold:
"children (r1 O r2) x = \<Union> (children r1 ` (children r2 x))"
by (auto simp:children_def)
lemma fbranch_compose:
assumes "fbranch r1"
and "fbranch r2"
shows "fbranch (r1 O r2)"
proof -
{ fix x
assume "x\<in>Range (r1 O r2)"
then obtain y z where h: "(y, z) \<in> r1" "(z, x) \<in> r2" by auto
have "finite (children (r1 O r2) x)"
proof(unfold children_compose_unfold, rule finite_Union)
show "finite (children r1 ` children r2 x)"
proof(rule finite_imageI)
from h(2) have "x \<in> Range r2" by auto
from assms(2)[unfolded fbranch_def, rule_format, OF this]
show "finite (children r2 x)" .
qed
next
fix M
assume "M \<in> children r1 ` children r2 x"
then obtain y where h1: "y \<in> children r2 x" "M = children r1 y" by auto
show "finite M"
proof(cases "children r1 y = {}")
case True
with h1(2) show ?thesis by auto
next
case False
hence "y \<in> Range r1" by (unfold children_def, auto)
from assms(1)[unfolded fbranch_def, rule_format, OF this, folded h1(2)]
show ?thesis .
qed
qed
} thus ?thesis by (unfold fbranch_def, auto)
qed
lemma finite_fbranchI:
assumes "finite r"
shows "fbranch r"
proof -
{ fix x
assume "x \<in>Range r"
have "finite (children r x)"
proof -
have "{y. (y, x) \<in> r} \<subseteq> Domain r" by (auto)
from rev_finite_subset[OF finite_Domain[OF assms] this]
have "finite {y. (y, x) \<in> r}" .
thus ?thesis by (unfold children_def, simp)
qed
} thus ?thesis by (auto simp:fbranch_def)
qed
lemma subset_fbranchI:
assumes "fbranch r1"
and "r2 \<subseteq> r1"
shows "fbranch r2"
proof -
{ fix x
assume "x \<in>Range r2"
with assms(2) have "x \<in> Range r1" by auto
from assms(1)[unfolded fbranch_def, rule_format, OF this]
have "finite (children r1 x)" .
hence "finite (children r2 x)"
proof(rule rev_finite_subset)
from assms(2)
show "children r2 x \<subseteq> children r1 x" by (auto simp:children_def)
qed
} thus ?thesis by (auto simp:fbranch_def)
qed
lemma children_subtree:
shows "children r x \<subseteq> subtree r x"
by (auto simp:children_def subtree_def)
lemma children_union_kept:
assumes "x \<notin> Range r'"
shows "children (r \<union> r') x = children r x"
using assms
by (auto simp:children_def)
end