Proof step 1 from proof get-from-01 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (x from m)[m0] ≠ x[m + m0], m + m0 < # x ⊦

Rule argument

Additional information

Proof step 2 from proof get-from-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  (x from m)[m0] ≠ x[m + m0], x ≠ [], m + m0 < # 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 get-from-01 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  a + x0 ≠ [], (a + x0 from m)[m0] ≠ (a + x0)[m + m0], m + m0 < #(a + x0), ∀ m, m0. ¬((x0 from m)[m0] ≠ x0[m + m0] ∧ x0 ≠ [] ∧ m + m0 < # x0) ⊦

Rule argument

Additional information

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

Proof step 4 from proof get-from-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (a ' + x0 from m)[m0] ≠ (a ' + x0)[m + m0], m + m0 ≤ # x0, ∀ m1, m2. (x0 from m1)[m2] = x0[m1 + m2] ∨ x0 = [] ∨ ¬ m1 + m2 < # 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 ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ a + x = a ' + x; (s) ⊦ a ' + x ≠ []; (s)

Proof step 5 from proof get-from-01 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  (a ' + x0 from m1 +1)[m0] ≠ (a ' + x0)[m1 +1 + m0], m1 +1 + m0 ≤ # x0, ∀ m1, m2. (x0 from m1)[m2] = x0[m1 + m2] ∨ x0 = [] ∨ ¬ m1 + m2 < # x0, ∀ a, x0, m0. ¬ ((a ' + x0 from m1)[m0] ≠ (a ' + x0)[m1 + m0] ∧ m1 + m0 ≤ # x0 ∧ (∀ m1, m2. (x0 from m1)[m2] = x0[m1 + m2] ∨ x0 = [] ∨ ¬ m1 + m2 < # x0)) ⊦

Rule argument

Additional information

Used simplifier rules: from-zero-l: ⊦ x from 0 = x; nat freely generated by 0, +1;; ⊦ 0 + m = m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 6 from proof get-from-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  (x0 from m1)[m0] ≠ x0[m1 + m0], x0 ≠ [], m1 + m0 < # x0, ∀ a, x, m. (a ' + x from m1)[m] = (a ' + x)[m1 + m] ∨ ¬ m1 + m ≤ # x ∨ ¬(∀ m1, m2. (x from m1)[m2] = x[m1 + m2] ∨ x = [] ∨ ¬ m1 + m2 < # x), ∀ m, m2. (x0 from m)[m2] = x0[m + m2] ∨ ¬ m + m2 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from 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 = m + 1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x)[n + 1] = x[n]; (s)

Proof step 7 from proof get-from-01 (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m1, m0