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