Proof step 1 from proof rev-subarray-split-last-swap (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0) ≠ subarray(ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]], m + 1, n + k + 2 + n0), k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Proof step 2 from proof rev-subarray-split-last-swap (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:

  3 + n0 + k + n + m ≤ # ar, k ≤ 1, #(subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0)) ≠ # subarray(ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]], m + 1, n + k + 2 + n0) ∨ (∃ n1. ¬ ( n1 < #(subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0)) → (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] = subarray(ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]], m + 1, n + k + 2 + n0)[n1])) ⊦

Rule argument

Additional information

Used terms: subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0), subarray (ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]], m + 1, n + k + 2 + n0)

Proof step 3 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 < n + (k + 2) + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ # subarray(ar, m, n) = n; (s) ⊦ # rev(x) = # x; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 5 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 6:

  (m + 1) + n1 = m + n + k + 2, (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

• Proof step 18:

  (m + 1) + n1 ≠ m + n + k + 2, (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used formulas: (m + 1) + n1 = m + n + k + 2

Proof step 6 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n + k + 1] ≠ ar[m + n + 1], k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) m < # ar ⊦ ar[m, a][m] = a; (s) n0 < n ⊦ ¬ m + (* n) ≤ (* n0); (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ # ar[n, a] = # ar; (s) lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem[natarray] n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 7 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ n < # x → (x + y)[n] = x[n]

• Proof step 9:

  (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n + k + 1] ≠ ar[m + n + 1], 3 + n0 + k + n + m ≤ # ar, k ≤ 1, ¬ n + k + 1 < # subarray(ar, m + 1, n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: subarray(ar, m + 1, n), rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0), n + k + 1

Proof step 8 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[k + 1] ≠ ar[m + n + 1], k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 < m; (s) ⊦ # subarray(ar, m, n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ m + n - m = n; (s) ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)

Proof step 10 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ n < # x → (x + y)[n] = x[n]

• Proof step 12:

  rev(subarray(ar, m + n + 1, k + 2))[k + 1] ≠ ar[m + n + 1], k + 1 < # rev(subarray(ar, m + n + 1, k + 2)), 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s) ⊦ m + n - n + n1 = m - n1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ # rev(x) = # x; (s) ⊦ # subarray(ar, m, n) = n; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: rev(subarray(ar, m + n + 1, k + 2)), subarray(ar, m + n + 3 + k, n0), k + 1

Proof step 11 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  rev(subarray(ar, m + n + 1, k + 2))[k + 1] ≠ ar[m + n + 1], k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 < m + n0 ↔ n < n0; (s) ⊦ # subarray(ar, m, n) = n; (s) ⊦ # rev(x) = # x; (s)

Proof step 13 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ n < # ar → rev(ar)[n] = ar[# ar - n + 1]

• Proof step 15:

  subarray(ar, m + n + 1, k + 2)[# subarray(ar, m + n + 1, k + 2) - (k + 1) + 1] ≠ ar[m + n + 1], k + 1 < # subarray(ar, m + n + 1, k + 2), 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: subarray(ar, m + n + 1, k + 2), k + 1

Proof step 14 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Lemma get-rev

Proof step 15 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  subarray(ar, m + n + 1, k + 2)[0] ≠ ar[m + n + 1], k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 < m + n0 ↔ n < n0; (s) ⊦ # subarray(ar, m, n) = n; (s) ⊦ m - m = 0; (s) ⊦ m + n - n + n1 = m - n1; (s)

Proof step 16 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ n + m ≤ # ar ∧ n0 < n → subarray(ar, m, n)[n0] = ar[m + 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 ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ m + n < m; (s) Used terms: ar, m + n + 1, k + 2, 0

Proof step 17 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 19 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ n < # x → (x + y)[n] = x[n]

• Proof step 21:

  (subarray(ar, m + 1, n) + rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar, ¬ n1 < # subarray(ar, m + 1, n) ⊦

• Proof step 58:

  subarray(ar, m + 1, n)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, n1 < # subarray(ar, m + 1, n), k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used terms: subarray(ar, m + 1, n), rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0), n1

Proof step 20 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 21 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n1 - n] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, n ≤ n1, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ m < m + n ↔ n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ # subarray(ar, m, n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)

Proof step 22 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 23:

  n1 = n2 + n, (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1, n ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n1, n, n2

Proof step 23 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m ≤ m + n; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 24 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 25:

  (m + 1) + n2 + n ≠ m + n + k + 2, (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: (m + 1) + n2 + n = m + n + k + 2

Proof step 25 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 26 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

  ⊦ n < # x → (x + y)[n] = x[n]

• Proof step 28:

  (rev(subarray(ar, m + n + 1, k + 2)) + subarray(ar, m + n + 3 + k, n0))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar, ¬ n2 < # rev(subarray(ar, m + n + 1, k + 2)) ⊦

• Proof step 40:

  rev(subarray(ar, m + n + 1, k + 2))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, n2 < # rev(subarray(ar, m + n + 1, k + 2)), k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used terms: rev(subarray(ar, m + n + 1, k + 2)), subarray(ar, m + n + 3 + k, n0), n2

Proof step 27 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 28 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  subarray(ar, m + n + 3 + k, n0)[n2 - k + 2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n0 ≠ 0, n2 ≠ k + 1, n2 < k + 2 + n0, k + 2 ≤ n2, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 < m + n ↔ n ≠ 0; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ # subarray(ar, m, n) = n; (s) ⊦ # rev(x) = # x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)

Proof step 29 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 30:

  n2 = n1 + k + 2, subarray(ar, m + n + 3 + k, n0)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n0 ≠ 0, n2 ≠ k + 1, n2 < k + 2 + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1, k + 2 ≤ n2 ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, k + 2, n1

Proof step 30 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][3 + m + n1 + k + n], n1 < n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m ≤ m + n; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 31 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 32:

  3 + m + n1 + k + n ≠ m + n + k + 2, ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][3 + m + n1 + k + n], n1 < n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: 3 + m + n1 + k + n = m + n + k + 2

Proof step 32 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][3 + m + n1 + k + n], n1 < n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 33 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

• Proof step 35:

  3 + m + n1 + k + n ≠ m + n + k + 2, ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][3 + m + n1 + k + n], n1 < n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: ar[m + n + 1, ar[m + n + k + 2]], m + n + k + 2, ar[m + n + 1], 3 + m + n1 + k + n

Proof step 34 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][3 + m + n1 + k + n], n1 < n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 36 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 37:

  3 + m + n1 + k + n ≠ m + n + 1, ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][3 + m + n1 + k + n], n1 < n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: 3 + m + n1 + k + n = m + n + 1

Proof step 37 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  ar[(m + n + 3 + k) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][3 + m + n1 + k + n], n1 < n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

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 = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 38 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) Used terms: ar, m + n + 1, ar[m + n + k + 2], 3 + m + n1 + k + n

Proof step 39 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 40 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  rev(subarray(ar, m + n + 1, k + 2))[n2] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2, n2 < k + 2 + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ # subarray(ar, m, n) = n; (s) ⊦ # rev(x) = # x; (s)

Proof step 41 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

  ⊦ n < # ar → rev(ar)[n] = ar[# ar - n + 1]

• Proof step 43:

  subarray(ar, m + n + 1, k + 2)[# subarray(ar, m + n + 1, k + 2) - n2 + 1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, n2 < k + 2, n2 < # subarray(ar, m + n + 1, k + 2), 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: subarray(ar, m + n + 1, k + 2), n2

Proof step 42 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Lemma get-rev

Proof step 43 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  subarray(ar, m + n + 1, k + 2)[k + 2 - n2 + 1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, n2 < k + 2, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

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

Proof step 44 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 45:

  k + 2 = n1 + n2 + 1, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2, n2 < k + 2 + n0, 3 + n0 + k + n + m ≤ # ar, k ≤ 1, n2 + 1 ≤ k + 2 ⊦

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ m + (* n) ≤ m0 + (* n0) ↔ m ≤ m0 + (n0 - n); (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → (n < m + 2 ↔ m + 1 = n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: k + 2, n2 + 1, n1

Proof step 45 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  n1 + n2 = k + 1, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, n2 < k + 2, k ≤ 1, 3 + n0 + k + n + m ≤ # ar, n2 ≤ k, n2 ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ 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) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) n < n0 ⊦ m + (* n) ≤ m0 + (* n0) ↔ m ≤ m0 + (n0 - n); (s) n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)

Proof step 46 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 47:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

• Proof step 48:

  n1 + n2 = k + 1, (m + 1) + n2 + n ≠ m + n + k + 2, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2, n2 < k + 2 + n0, n2 ≤ 1, n2 ≤ k, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: ar[m + n + 1, ar[m + n + k + 2]], m + n + k + 2, ar[m + n + 1], (m + 1) + n2 + n

Proof step 47 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 48 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  n1 + n2 = k + 1, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2 + n0, n2 < k + 2, k ≤ 1, 3 + n0 + k + n + m ≤ # ar, n2 ≤ k, n2 ≤ 1 ⊦

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 n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 49 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 50:

  n1 + n2 = k + 1, (m + 1) + n2 + n = m + n + 1, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2, n2 < k + 2 + n0, n2 ≤ 1, n2 ≤ k, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

• Proof step 53:

  n1 + n2 = k + 1, (m + 1) + n2 + n ≠ m + n + 1, subarray(ar, m + n + 1, k + 2)[n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n2 + n], n2 ≠ k + 1, n2 < k + 2, n2 < k + 2 + n0, n2 ≤ 1, n2 ≤ k, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used formulas: (m + 1) + n2 + n = m + n + 1

Proof step 50 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  subarray(ar, m + n + 1, k + 2)[k + 1] ≠ ar[m + n + k + 2], k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ 0 < m ↔ m ≠ 0; (s) m < # ar ⊦ ar[m, a][m] = a; (s) n0 < n ⊦ ¬ m + (* n) ≤ (* n0); (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m + 1 ≠ 0; (s) ⊦ n + 0 = n; (s) lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem[natarray] ⊦ m = m + n ↔ n = 0; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 51 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 52:

  ⊦ n + m ≤ # ar ∧ n0 < n → subarray(ar, m, n)[n0] = ar[m + 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; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ m + n < m; (s) Used terms: ar, m + n + 1, k + 2, k + 1

Proof step 52 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 53 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

  ar[2 + m + n] ≠ ar[m + n + 1, ar[3 + m + n]][2 + m + n], 4 + n0 + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0); (ws) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) ⊦ m + n = m + n0 ↔ n = n0; (s) lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem[natarray] ⊦ m = m + n ↔ n = 0; (s)

Proof step 54 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 55:

  2 + m + n ≠ m + n + 1, ar[2 + m + n] ≠ ar[m + n + 1, ar[3 + m + n]][2 + m + n], 4 + n0 + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: 2 + m + n = m + n + 1

Proof step 55 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

  ar[2 + m + n] ≠ ar[m + n + 1, ar[3 + m + n]][2 + m + n], 4 + n0 + n + m ≤ # ar ⊦

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 = m + n0 ↔ n = n0; (s)

Proof step 56 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 57:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) Used terms: ar, m + n + 1, ar[3 + m + n], 2 + m + n

Proof step 57 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 58 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  ar[(m + 1) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][m + n + k + 2, ar[m + n + 1]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n, n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) ⊦ # subarray(ar, m, n) = n; (s)

Proof step 59 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 60:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

• Proof step 61:

  (m + 1) + n1 ≠ m + n + k + 2, ar[(m + 1) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, n1 < n, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: ar[m + n + 1, ar[m + n + k + 2]], m + n + k + 2, ar[m + n + 1], (m + 1) + n1

Proof step 60 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma

Proof step 61 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 62:

  ar[(m + 1) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n, n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 62 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 63:

  (m + 1) + n1 ≠ m + n + 1, ar[(m + 1) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n1], n1 ≠ n + k + 1, n1 < n + (k + 2) + n0, n1 < n, 3 + n0 + k + n + m ≤ # ar, k ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem[natarray] lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ m < m + n ↔ n ≠ 0; (s) n < n0 ⊦ (* n) ≠ m0 + (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → ¬ n < m; (s) Used formulas: (m + 1) + n1 = m + n + 1

Proof step 63 from proof rev-subarray-split-last-swap (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 64:

  ar[(m + 1) + n1] ≠ ar[m + n + 1, ar[m + n + k + 2]][(m + 1) + n1], n1 ≠ n + k + 1, n1 ≠ n, n1 < n, n1 < n + (k + 2) + n0, k ≤ 1, 3 + n0 + k + n + m ≤ # ar ⊦

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 = m + n0 ↔ n = n0; (s)

Proof step 64 from proof rev-subarray-split-last-swap (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 65:

  ⊦ m ≠ n → ar[n, a][m] = ar[m]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) Used terms: ar, m + n + 1, ar[m + n + k + 2], (m + 1) + n1

Proof step 65 from proof rev-subarray-split-last-swap (previous step) (back to tree)

A spec-lemma