substitution lemma in separate file
authorChristian Urban <urbanc@in.tum.de>
Fri, 21 Jan 2011 22:02:34 +0100
changeset 2690 f325eefe803e
parent 2689 ddc05a611005
child 2691 abb6c3ac2df2
substitution lemma in separate file
Tutorial/Tutorial1.thy
Tutorial/Tutorial3.thy
--- a/Tutorial/Tutorial1.thy	Fri Jan 21 21:58:51 2011 +0100
+++ b/Tutorial/Tutorial1.thy	Fri Jan 21 22:02:34 2011 +0100
@@ -820,146 +820,6 @@
   finally show "((E # Es1) @ Es2)\<down> = Es2\<down> \<odot> (E # Es1)\<down>" by simp
 qed
 
-text {******************************************************************
-  
-  Formalising Barendregt's Proof of the Substitution Lemma
-  --------------------------------------------------------
-
-  Barendregt's proof needs in the variable case a case distinction.
-  One way to do this in Isar is to use blocks. A block is some sequent
-  or reasoning steps enclosed in curly braces
-
-  { \<dots>
-
-    have "statement"
-  }
-
-  Such a block can contain local assumptions like
-
-  { assume "A"
-    assume "B"
-    \<dots>
-    have "C" by \<dots>
-  }
-
-  Where "C" is the last have-statement in this block. The behaviour 
-  of such a block to the 'outside' is the implication
-
-   \<lbrakk>A; B\<rbrakk> \<Longrightarrow> "C" 
-
-  Now if we want to prove a property "smth" using the case-distinctions
-  P1, P2 and P3 then we can use the following reasoning:
-
-    { assume "P1"
-      \<dots>
-      have "smth"
-    }
-    moreover
-    { assume "P2"
-      \<dots>
-      have "smth"
-    }
-    moreover
-    { assume "P3"
-      \<dots>
-      have "smth"
-    }
-    ultimately have "smth" by blast
-
-  The blocks establish the implications
-
-    P1 \<Longrightarrow> smth
-    P2 \<Longrightarrow> smth
-    P3 \<Longrightarrow> smth
-
-  If we know that P1, P2 and P3 cover all the cases, that is P1 \<or> P2 \<or> P3 is
-  true, then we have 'ultimately' established the property "smth" 
-  
-*}
-
-section {* EXERCISE 7 *}
-
-text {*
-  Fill in the cases 1.2 and 1.3 and the equational reasoning 
-  in the lambda-case.
-*}
-
-lemma forget:
-  shows "atom x \<sharp> t \<Longrightarrow> t[x ::= s] = t"
-by (nominal_induct t avoiding: x s rule: lam.strong_induct)
-   (auto simp add: lam.fresh fresh_at_base)
-
-lemma fresh_fact:
-  assumes a: "atom z \<sharp> s"
-  and b: "z = y \<or> atom z \<sharp> t"
-  shows "atom z \<sharp> t[y ::= s]"
-using a b
-by (nominal_induct t avoiding: z y s rule: lam.strong_induct)
-   (auto simp add: lam.fresh fresh_at_base)
-
-
-lemma 
-  assumes a: "x \<noteq> y"
-  and     b: "atom x \<sharp> L"
-  shows "M[x::=N][y::=L] = M[y::=L][x::=N[y::=L]]"
-using a b
-proof (nominal_induct M avoiding: x y N L rule: lam.strong_induct)
-  case (Var z)
-  have a1: "x \<noteq> y" by fact
-  have a2: "atom x \<sharp> L" by fact
-  show "Var z[x::=N][y::=L] = Var z[y::=L][x::=N[y::=L]]" (is "?LHS = ?RHS")
-  proof -
-    { -- {* Case 1.1 *}
-      assume c1: "z = x"
-      have "(1)": "?LHS = N[y::=L]" using c1 by simp
-      have "(2)": "?RHS = N[y::=L]" using c1 a1 by simp
-      have "?LHS = ?RHS" using "(1)" "(2)" by simp
-    }
-    moreover 
-    { -- {* Case 1.2 *}
-      assume c2: "z = y" "z \<noteq> x" 
-      
-      have "?LHS = ?RHS" sorry
-    }
-    moreover 
-    { -- {* Case 1.3 *}
-      assume c3: "z \<noteq> x" "z \<noteq> y"
-      
-      have "?LHS = ?RHS" sorry
-    }
-    ultimately show "?LHS = ?RHS" by blast
-  qed
-next
-  case (Lam z M1) -- {* case 2: lambdas *}
-  have ih: "\<lbrakk>x \<noteq> y; atom x \<sharp> L\<rbrakk> \<Longrightarrow> M1[x::=N][y::=L] = M1[y::=L][x::=N[y::=L]]" by fact
-  have a1: "x \<noteq> y" by fact
-  have a2: "atom x \<sharp> L" by fact
-  have fs: "atom z \<sharp> x" "atom z \<sharp> y" "atom z \<sharp> N" "atom z \<sharp> L" by fact+
-  then have b: "atom z \<sharp> N[y::=L]" by (simp add: fresh_fact)
-  show "(Lam [z].M1)[x::=N][y::=L] = (Lam [z].M1)[y::=L][x::=N[y::=L]]" (is "?LHS=?RHS") 
-  proof - 
-    have "?LHS = \<dots>" sorry
-
-    also have "\<dots> = ?RHS" sorry
-    finally show "?LHS = ?RHS" by simp
-  qed
-next
-  case (App M1 M2) -- {* case 3: applications *}
-  then show "(App M1 M2)[x::=N][y::=L] = (App M1 M2)[y::=L][x::=N[y::=L]]" by simp
-qed
-
-text {* 
-  Again the strong induction principle enables Isabelle to find
-  the proof of the substitution lemma automatically. 
-*}
-
-lemma substitution_lemma_version:  
-  assumes asm: "x \<noteq> y" "atom x \<sharp> L"
-  shows "M[x::=N][y::=L] = M[y::=L][x::=N[y::=L]]"
-  using asm 
-by (nominal_induct M avoiding: x y N L rule: lam.strong_induct)
-   (auto simp add: fresh_fact forget)
-
 
 
 end  
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/Tutorial/Tutorial3.thy	Fri Jan 21 22:02:34 2011 +0100
@@ -0,0 +1,144 @@
+theory Tutorial3
+imports Lambda
+begin
+
+section {* Formalising Barendregt's Proof of the Substitution Lemma *}
+
+text {*
+  Barendregt's proof needs in the variable case a case distinction.
+  One way to do this in Isar is to use blocks. A block is some sequent
+  or reasoning steps enclosed in curly braces
+
+  { \<dots>
+
+    have "statement"
+  }
+
+  Such a block can contain local assumptions like
+
+  { assume "A"
+    assume "B"
+    \<dots>
+    have "C" by \<dots>
+  }
+
+  Where "C" is the last have-statement in this block. The behaviour 
+  of such a block to the 'outside' is the implication
+
+   \<lbrakk>A; B\<rbrakk> \<Longrightarrow> "C" 
+
+  Now if we want to prove a property "smth" using the case-distinctions
+  P1, P2 and P3 then we can use the following reasoning:
+
+    { assume "P1"
+      \<dots>
+      have "smth"
+    }
+    moreover
+    { assume "P2"
+      \<dots>
+      have "smth"
+    }
+    moreover
+    { assume "P3"
+      \<dots>
+      have "smth"
+    }
+    ultimately have "smth" by blast
+
+  The blocks establish the implications
+
+    P1 \<Longrightarrow> smth
+    P2 \<Longrightarrow> smth
+    P3 \<Longrightarrow> smth
+
+  If we know that P1, P2 and P3 cover all the cases, that is P1 \<or> P2 \<or> P3 is
+  true, then we have 'ultimately' established the property "smth" 
+  
+*}
+
+section {* EXERCISE 7 *}
+
+text {*
+  Fill in the cases 1.2 and 1.3 and the equational reasoning 
+  in the lambda-case.
+*}
+
+lemma forget:
+  shows "atom x \<sharp> t \<Longrightarrow> t[x ::= s] = t"
+by (nominal_induct t avoiding: x s rule: lam.strong_induct)
+   (auto simp add: lam.fresh fresh_at_base)
+
+lemma fresh_fact:
+  assumes a: "atom z \<sharp> s"
+  and b: "z = y \<or> atom z \<sharp> t"
+  shows "atom z \<sharp> t[y ::= s]"
+using a b
+by (nominal_induct t avoiding: z y s rule: lam.strong_induct)
+   (auto simp add: lam.fresh fresh_at_base)
+
+
+lemma 
+  assumes a: "x \<noteq> y"
+  and     b: "atom x \<sharp> L"
+  shows "M[x::=N][y::=L] = M[y::=L][x::=N[y::=L]]"
+using a b
+proof (nominal_induct M avoiding: x y N L rule: lam.strong_induct)
+  case (Var z)
+  have a1: "x \<noteq> y" by fact
+  have a2: "atom x \<sharp> L" by fact
+  show "Var z[x::=N][y::=L] = Var z[y::=L][x::=N[y::=L]]" (is "?LHS = ?RHS")
+  proof -
+    { -- {* Case 1.1 *}
+      assume c1: "z = x"
+      have "(1)": "?LHS = N[y::=L]" using c1 by simp
+      have "(2)": "?RHS = N[y::=L]" using c1 a1 by simp
+      have "?LHS = ?RHS" using "(1)" "(2)" by simp
+    }
+    moreover 
+    { -- {* Case 1.2 *}
+      assume c2: "z = y" "z \<noteq> x" 
+      
+      have "?LHS = ?RHS" sorry
+    }
+    moreover 
+    { -- {* Case 1.3 *}
+      assume c3: "z \<noteq> x" "z \<noteq> y"
+      
+      have "?LHS = ?RHS" sorry
+    }
+    ultimately show "?LHS = ?RHS" by blast
+  qed
+next
+  case (Lam z M1) -- {* case 2: lambdas *}
+  have ih: "\<lbrakk>x \<noteq> y; atom x \<sharp> L\<rbrakk> \<Longrightarrow> M1[x::=N][y::=L] = M1[y::=L][x::=N[y::=L]]" by fact
+  have a1: "x \<noteq> y" by fact
+  have a2: "atom x \<sharp> L" by fact
+  have fs: "atom z \<sharp> x" "atom z \<sharp> y" "atom z \<sharp> N" "atom z \<sharp> L" by fact+
+  then have b: "atom z \<sharp> N[y::=L]" by (simp add: fresh_fact)
+  show "(Lam [z].M1)[x::=N][y::=L] = (Lam [z].M1)[y::=L][x::=N[y::=L]]" (is "?LHS=?RHS") 
+  proof - 
+    have "?LHS = \<dots>" sorry
+
+    also have "\<dots> = ?RHS" sorry
+    finally show "?LHS = ?RHS" by simp
+  qed
+next
+  case (App M1 M2) -- {* case 3: applications *}
+  then show "(App M1 M2)[x::=N][y::=L] = (App M1 M2)[y::=L][x::=N[y::=L]]" by simp
+qed
+
+text {* 
+  Again the strong induction principle enables Isabelle to find
+  the proof of the substitution lemma automatically. 
+*}
+
+lemma substitution_lemma_version:  
+  assumes asm: "x \<noteq> y" "atom x \<sharp> L"
+  shows "M[x::=N][y::=L] = M[y::=L][x::=N[y::=L]]"
+  using asm 
+by (nominal_induct M avoiding: x y N L rule: lam.strong_induct)
+   (auto simp add: fresh_fact forget)
+
+
+end