Proof step 1 from proof map-get (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (x map f)[n] ≠ f(x[n]), n < # x ⊦

Rule argument

Additional information

Proof step 2 from proof map-get (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  (x map f)[n] ≠ f(x[n]), x ≠ [], n < # x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof map-get (previous step) (back to tree)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

• Proof step 4:

  a + x0 ≠ [], ((a + x0) map f)[n] ≠ f((a + x0)[n]), n < #(a + x0), ∀ f, n. ¬((x0 map f)[n] ≠ f(x0[n]) ∧ x0 ≠ [] ∧ n < # x0) ⊦

Rule argument

Additional information

Used simplifier rules: list('a) freely generated by [], +;;

Proof step 4 from proof map-get (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (f(a) ' + (x0 map f))[n] ≠ f((a ' + x0)[n]), n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a + x = a ' + x; (s) ⊦ (a ' + x) map f = f(a) + (x map f); (s) ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

Proof step 5 from proof map-get (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 6:

  (f(a) ' + (x0 map f))[n] ≠ f((a ' + x0)[n]), x0 ≠ [], n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: one-03: ⊦ a '[0] = a; ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # [] = 0; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ [] map f = []; (s) Used formulas: x0 = []

Proof step 6 from proof map-get (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  (f(a) ' + (x0 map f))[n] ≠ f((a ' + x0)[n]), x0 ≠ [], n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Proof step 7 from proof map-get (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]

• Proof step 9:

  a ' + x0 ≠ [], (f(a) ' + (x0 map f))[n] ≠ f((a ' + x0).tail[n - 1]), n ≠ 0, x0 ≠ [], n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: get-zero: ⊦ (a ' + x)[0] = a; ⊦ x + y = [] ↔ x = [] ∧ y = []; (s) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + x0, n

Proof step 8 from proof map-get (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof map-get (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  (f(a) ' + (x0 map f))[n] ≠ f(x0[n - 1]), n ≠ 0, x0 ≠ [], n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

Proof step 10 from proof map-get (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 11:

  n = m + 1, (f(a) ' + (x0 map f))[n] ≠ f(x0[m]), n ≠ 0, x0 ≠ [], n ≤ # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m

Proof step 11 from proof map-get (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  (x0 map f)[m] ≠ f(x0[m]), x0 ≠ [], m < # x0, ∀ f0, n0. (x0 map f0)[n0] = f0(x0[n0]) ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n]; ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m + 1 ≠ 0; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 12 from proof map-get (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: f, m