theory Tutorial3s
imports Lambda
begin

section {* Formalising Barendregt's Proof of the Substitution Lemma *}

text {*
  The substitution lemma is another theorem where the variable
  convention plays a crucial role.

  Barendregt's proof of this lemma needs in the variable case a 
  case distinction. One way to do this in Isar is to use blocks. 
  A block consist of some assumptions and reasoning steps 
  enclosed in curly braces, like

  { \<dots>
    have "statement"
    have "last_statement_in_the_block"
  }

  Such a block may 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

   A \<Longrightarrow> B \<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 
  holds, then we have 'ultimately' established the property "smth" 
  
*}

subsection {* Two preliminary facts *}

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)



section {* EXERCISE 10 *}

text {*
  Fill in the cases 1.2 and 1.3 and the equational reasoning 
  in the lambda-case.
*}

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 "(1)": "?LHS = L" using c2 by simp
      have "(2)": "?RHS = L[x::=N[y::=L]]" using c2 by simp
      have "(3)": "L[x::=N[y::=L]] = L" using a2 forget by simp
      have "?LHS = ?RHS" using "(1)" "(2)" "(3)" by simp
    }
    moreover 
    { -- {* Case 1.3 *}
      assume c3: "z \<noteq> x" "z \<noteq> y"
      have "(1)": "?LHS = Var z" using c3 by simp
      have "(2)": "?RHS = Var z" using c3 by simp
      have "?LHS = ?RHS" using "(1)" "(2)" by simp
    }
    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 = Lam [z].(M1[x ::= N][y ::= L])" using fs by simp
    also have "\<dots> = Lam [z].(M1[y ::= L][x ::= N[y ::= L]])" using ih a1 a2 by simp
    also have "\<dots> = (Lam [z].(M1[y ::= L]))[x ::= N[y ::= L]]" using b fs by simp
    also have "\<dots> = ?RHS" using fs by simp
    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 completely 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)

subsection {* MINI EXERCISE *}

text {*
  Compare and contrast Barendregt's reasoning and the 
  formalised proofs.
*}

end