Proof step 1 from proof cut (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ar ≠ [] ⊦ ∃ a, ar0. ar = a + ar0

Rule argument

Additional information

Proof step 2 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ar ≠ [] ⊦ ∃ a, ar0. ar = a + ar0

Rule argument

Additional information

Proof step 3 from proof cut (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 4:

  ar ≠ ar.first + ar.rest, ar ≠ [] ⊦

Rule argument

Additional information

Used terms: ar.first, ar.rest

Proof step 4 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

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

• Proof step 6:

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

Rule argument

Additional information

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

Proof step 5 from proof cut (previous step) (back to tree)

A spec-lemma

Proof step 6 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

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

Rule argument

Additional information

Proof step 7 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ ar.first = ar[0]

• Proof step 9:

  ar ≠ [], # ar ≠ #(ar[0] + ar.rest) ∨ ¬(n < # ar → ar[n] = (ar[0] + ar.rest)[n]) ⊦

Rule argument

Additional information

Used terms: ar

Proof step 8 from proof cut (previous step) (back to tree)

Lemma first

Proof step 9 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

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

Rule argument

Additional information

Used terms: ar

Proof step 10 from proof cut (previous step) (back to tree)

Lemma rest

Proof step 11 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ a + ar = copy(ar, 0, mkarray(# ar + 1), 1, # ar)[0 := a]

• Proof step 13:

  ar ≠ [], # ar ≠ # copy(subarray(ar, 1, # ar - 1), 0, mkarray(# subarray(ar, 1, # ar - 1) + 1), 1, # subarray(ar, 1, # ar - 1))[0 := ar[0]] ∨ ¬ ( n < # ar → ar[n] = (copy(subarray(ar, 1, # ar - 1), 0, mkarray(# subarray(ar, 1, # ar - 1) + 1), 1, # subarray(ar, 1, # ar - 1))[0 := ar[0]] )[n ]) ⊦

Rule argument

Additional information

Used terms: subarray(ar, 1, # ar - 1), ar[0]

Proof step 12 from proof cut (previous step) (back to tree)

Lemma cons

Proof step 13 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  ar ≠ [], # ar ≠ (# ar - 1) + 1 ∨ ¬(n < # ar → ar[n] = (copy(ar, 1, mkarray((# ar - 1) + 1), 1, # ar - 1)[0 := ar[0]])[n]) ⊦

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 n0 + n1 ≤ # ar0, m + m1 ≤ n1 ⊦ copy(subarray(ar0, n0, n1), m, ar1, m0, m1) = copy(ar0, n0 + m, ar1, m0, m1); (s) ⊦ ¬ n + (m - n) < m; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s) ⊦ m < n → n ≠ 0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ n + 0 = n; (s) ⊦ m ≤ m; (s) ⊦ # subarray(ar, m, n) = n; (s) ⊦ # mkarray(n) = n; (s) ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s) ⊦ # ar[n := a] = # ar; (s)

Proof step 14 from proof cut (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 15:

  # ar = 0, ar ≠ [], # ar ≠ (# ar - 1) + 1 ∨ ¬(n < # ar → ar[n] = (copy(ar, 1, mkarray((# ar - 1) + 1), 1, # ar - 1)[0 := ar[0]])[n]) ⊦

• Proof step 21:

  # ar = n0 + 1, # ar ≠ 0, ar ≠ [], # ar ≠ n0 + 1 ∨ ¬(n < # ar → ar[n] = (copy(ar, 1, mkarray(n0 + 1), 1, n0)[0 := ar[0]])[n]) ⊦

Rule argument

Additional information

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

Proof step 15 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # ar = 0, 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 + 1 ≠ 0; (s)

Proof step 16 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

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

• Proof step 18:

  # ar = 0, # ar ≠ # [] ∨ (∃ n. ¬(n < # ar → ar[n] = [][n])) ⊦

Rule argument

Additional information

Used terms: [], ar

Proof step 17 from proof cut (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  # ar = 0, 0 ≠ # [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 19 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ [] = mkarray(0)

Rule argument

Additional information

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

Proof step 20 from proof cut (previous step) (back to tree)

Lemma nil

Proof step 21 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # ar = n0 + 1, ar[n] ≠ (copy(ar, 1, mkarray(n0 + 1), 1, n0)[0 := ar[0]])[n], ar ≠ [], n ≤ n0 ⊦

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) ⊦ m + 1 ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s)

Proof step 22 from proof cut (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 23:

  # ar = n0 + 1, ar[n] ≠ (copy(ar, 1, mkarray(n0 + 1), 1, n0)[0 := ar[0]])[n], ar ≠ [], n ≠ 0, n ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s) ⊦ # mkarray(n) = n; (s) ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ m + 1 ≠ 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < # ar ⊦ (ar[m := a])[m] = a; (s) Used formulas: n = 0

Proof step 23 from proof cut (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # ar = n0 + 1, ar[n] ≠ copy(ar, 1, mkarray(n0 + 1), 1, n0)[n], ar ≠ [], n ≠ 0, n ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ n → (ar[n := a])[m] = ar[m]; (s)

Proof step 24 from proof cut (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])

Rule argument

Additional information

Used simplifier rules: ⊦ m < 1 ↔ m = 0; (s) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ # mkarray(n) = n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) Used terms: 1, 1, mkarray(n0 + 1), n0, n, ar

Proof step 25 from proof cut (previous step) (back to tree)

A spec-lemma