Proof step 1 from proof case-list (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  x ≠ [], x ≠ x.first + x.rest ⊦

Rule argument

Additional information

Proof step 2 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])

• Proof step 4:

  x ≠ [], # x ≠ #(x.first + x.rest) ∨ (∃ n. ¬(n < # x → x[n] = (x.first + x.rest)[n])) ⊦

Rule argument

Additional information

Used terms: x, x.first + x.rest

Proof step 3 from proof case-list (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  x ≠ [], # x ≠ #(x.first + x.rest) ∨ ¬(n < # x → x[n] = (x.first + x.rest)[n]) ⊦

Rule argument

Additional information

Proof step 5 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ #(a + x) = (# x)+1

• Proof step 7:

  x ≠ [], # x ≠ (# x.rest)+1 ∨ ¬(n < # x → x[n] = (x.first + x.rest)[n]) ⊦

Rule argument

Additional information

Used terms: x.first, x.rest

Proof step 6 from proof case-list (previous step) (back to tree)

Lemma size-add

Proof step 7 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  x ≠ [], # x ≠ # x.rest + 1 ∨ ¬(n < # x → x[n] = (x.first + x.rest)[n]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)

Proof step 8 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ ar.rest = subarray(ar, 1, # ar - 1)

• Proof step 10:

  x ≠ [], # x ≠ # subarray(x, 1, # x - 1) + 1 ∨ ¬(n < # x → x[n] = (x.first + subarray(x, 1, # x - 1))[n]) ⊦

Rule argument

Additional information

Used terms: x

Proof step 9 from proof case-list (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  x ≠ [], # x ≠ (# x - 1) + 1 ∨ ¬(n < # x → x[n] = (x.first + subarray(x, 1, # x - 1))[n]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)

Proof step 11 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ ar.first = ar[0]

• Proof step 13:

  x ≠ [], # x ≠ (# x - 1) + 1 ∨ ¬(n < # x → x[n] = (x[0] + subarray(x, 1, # x - 1))[n]) ⊦

Rule argument

Additional information

Used terms: x

Proof step 12 from proof case-list (previous step) (back to tree)

A spec-lemma

Proof step 13 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  n ≤ m ⊦ (m - n) + n = m

• Proof step 15:

  x ≠ [], ¬ 1 ≤ # x, # x ≠ (# x - 1) + 1 ∨ ¬(n < # x → x[n] = (x[0] + subarray(x, 1, # x - 1))[n]) ⊦

• Proof step 20:

  x ≠ [], 1 ≤ # x, # x ≠ # x ∨ ¬(n < # x → x[n] = (x[0] + subarray(x, 1, # x - 1))[n]) ⊦

Rule argument

Additional information

Used terms: # x, 1

Proof step 14 from proof case-list (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # x = 0, x ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 16 from proof case-list (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 17:

  # x = 0, x = mkarray(n), x ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: array('a) generated by mkarray, [ := ];; ⊦ # ar[n := a] = # ar; (s) ⊦ ar[n := a] = ar ↔ ¬ n < # ar ∨ ar[n] = a; (s) ⊦ m < n → n ≠ 0; (s)

Proof step 17 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  mkarray(0) ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # mkarray(n) = n; (s)

Proof step 18 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ [] = mkarray(0)

Rule argument

Additional information

Used terms:

Proof step 19 from proof case-list (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  x[n] ≠ (x[0] + subarray(x, 1, # x - 1))[n], x ≠ [], n < # x, 1 ≤ # x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 21 from proof case-list (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 22:

  # x = n0 + 1, # x ≠ 0, x[n] ≠ (x[0] + subarray(x, 1, n0))[n], x ≠ [], n < # x, 1 ≤ # x ⊦

Rule argument

Additional information

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

Proof step 22 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # x = n0 + 1, x[n] ≠ (x[0] + subarray(x, 1, n0))[n], x ≠ [], n ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ m + n; (s) ⊦ m + 1 ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s)

Proof step 23 from proof case-list (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 24:

  # x = n0 + 1, n = 0, x[n] ≠ (x[0] + subarray(x, 1, n0))[n], x ≠ [], n ≤ n0 ⊦

• Proof step 27:

  # x = n0 + 1, n = n1 +1, x[n] ≠ (x[0] + subarray(x, 1, n0))[n], x ≠ [], n ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 24 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # x = n0 + 1, x[0] ≠ (x[0] + subarray(x, 1, n0))[0], x ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)

Proof step 25 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ (a + x)[0] = a

Rule argument

Additional information

Used terms: x[0], subarray(x, 1, n0)

Proof step 26 from proof case-list (previous step) (back to tree)

Lemma get-zero

Proof step 27 from proof case-list (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # x = n0 + 1, x[n1 + 1] ≠ (x[0] + subarray(x, 1, n0))[n1 + 1], x ≠ [], n1 < n0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (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)

Proof step 28 from proof case-list (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ (a + x)[n + 1] = x[n]

Rule argument

Additional information

Used simplifier rules: ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) c: ⊦ m + n = n + m; from specification nat-basic Used terms: x[0], subarray(x, 1, n0), n1

Proof step 29 from proof case-list (previous step) (back to tree)

Lemma get-succ