Proof step 1 from proof trans-List (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  x < x0, x0 < x1, ¬ x < x1 ⊦

Rule argument

Additional information

Proof step 2 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ ar0 < ar1 ↔ # ar0 < # ar1 ∧ ar0 = subarray(ar1, # ar1 - # ar0, # ar0)

• Proof step 4:

  x0 < x1, x < x0, ¬ # x < # x1 ∨ x ≠ subarray(x1, # x1 - # x, # x) ⊦

Rule argument

Additional information

Used terms: x, x1

Proof step 3 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

  ⊦ ar0 < ar1 ↔ # ar0 < # ar1 ∧ ar0 = subarray(ar1, # ar1 - # ar0, # ar0)

• Proof step 6:

  x = subarray(x0, # x0 - # x, # x), # x < # x0, x0 < x1, ¬ # x < # x1 ∨ x ≠ subarray(x1, # x1 - # x, # x) ⊦

Rule argument

Additional information

Used terms: x, x0

Proof step 5 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 6 from proof trans-List (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 7:

  # x0 = n0 + # x, x = subarray(x0, n0, # x), x0 < x1, # x < # x0, # x ≤ # x0, ¬ # x < # x1 ∨ x ≠ subarray(x1, # x1 - # x, # x) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, # x, n0

Proof step 7 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # x0 = n0 + # x, x = subarray(x0, n0, # x), n0 ≠ 0, x0 < x1, # x < # x0, ¬ # x < # x1 ∨ x ≠ subarray(x1, # x1 - # x, # 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 < m + n ↔ n ≠ 0; (s) ⊦ m < n → n ≠ 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 → m ≤ n; (s)

Proof step 8 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ ar0 < ar1 ↔ # ar0 < # ar1 ∧ ar0 = subarray(ar1, # ar1 - # ar0, # ar0)

• Proof step 10:

  # x0 = n0 + # x, x = subarray(x0, n0, # x), x0 = subarray(x1, # x1 - # x0, # x0), n0 ≠ 0, # x < # x0, # x0 < # x1, ¬ # x < # x1 ∨ x ≠ subarray(x1, # x1 - # x, # x) ⊦

Rule argument

Additional information

Used terms: x0, x1

Proof step 9 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # x0 = n0 + # x, x = subarray(x0, n0, # x), x0 = subarray(x1, # x1 - # x0, # x0), n0 ≠ 0, x ≠ subarray(x1, # x1 - # x, # x), # x < # x1, # x < # x0, # x0 < # x1, # x + 1 < # x1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic

Proof step 11 from proof trans-List (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 12:

  # x0 = n0 + # x, # x1 = n1 + # x0, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), n0 ≠ 0, x ≠ subarray(x1, # x1 - # x, # x), # x + 1 < # x1, # x0 < # x1, # x < # x0, # x < # x1, # x0 ≤ # x1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x1, # x0, n

Proof step 12 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # x0 = n0 + # x, # x1 = n1 + # x0, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), n0 ≠ 0, n1 ≠ 0, x ≠ subarray(x1, # x1 - # x, # x), # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1 ⊦

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) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 13 from proof trans-List (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), n0 ≠ 0, n1 ≠ 0, x ≠ subarray(x1, n2, # x), # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, # x ≤ # x1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x1, # x, n

Proof step 14 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), n0 ≠ 0, n1 ≠ 0, n2 ≠ 1, n2 ≠ 0, x ≠ subarray(x1, n2, # x), # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1 ⊦

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) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 15 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), n0 ≠ 0, n1 ≠ 0, n2 ≠ 1, n2 ≠ 0, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, # x ≠ # subarray(x1, n2, # x) ∨ (∃ n. ¬(n < # x → x(n) = subarray(x1, n2, # x)(n))) ⊦

Rule argument

Additional information

Used terms: x, subarray(x1, n2, # x)

Proof step 16 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x = subarray(x0, n0, # x), x0 = subarray(x1, n1, # x0), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (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 = m + 1; (s) ⊦ m + n < m0 → m + n < m0 + n0; (s) trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic ⊦ 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 18 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

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

• Proof step 20:

  # x = # subarray(x0, n0, # x), # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x0 = subarray(x1, n1, # x0), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < # x1, n < # x1, n < # x0, # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, n < # x, 2 + n < # x1, ∀ n. n < # x → x(n) = subarray(x0, n0, # x)(n) ⊦

Rule argument

Additional information

Used terms: x, subarray(x0, n0, # x)

Proof step 19 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x0 = subarray(x1, n1, # x0), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n0 < n → subarray(ar, m, n)(n0) = ar(m + n0); (s) ⊦ # subarray(ar, m, n) = n; (s)

Proof step 21 from proof trans-List (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

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

• Proof step 23:

  # x0 = # subarray(x1, n1, # x0), # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < # x1, n < # x1, n < # x0, # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, n < # x, 2 + n < # x1, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3), ∀ n. n < # x0 → x0(n) = subarray(x1, n1, # x0)(n) ⊦

Rule argument

Additional information

Used terms: x0, subarray(x1, n1, # x0)

Proof step 22 from proof trans-List (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n0 < n → subarray(ar, m, n)(n0) = ar(m + n0); (s) ⊦ # subarray(ar, m, n) = n; (s)

Proof step 24 from proof trans-List (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 25:

  # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < # x1, n < # x1, n < # x0, # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, n < # x, 2 + n < # x1, n < # x → x(n) = x0(n0 + n), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3), ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

Rule argument

Additional information

Used terms: n

Proof step 25 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x(n) = x0(n0 + n), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3) ⊦

Rule argument

Additional information

Proof step 26 from proof trans-List (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 27:

  # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x(n) = x0(n0 + n), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < # x1, n < # x1, n < # x0, # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, n < # x, 2 + n < # x1, n0 + n < # x0 → x0(n0 + n) = x1(n1 + n0 + n), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3), ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

Rule argument

Additional information

Used terms: n0 + n

Proof step 27 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x(n) = x0(n0 + n), x(n) = x1(n1 + n0 + n), x(n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3) ⊦

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 = m0 + n0 ∧ n < n0 → m0 + n < m; (s)

Proof step 28 from proof trans-List (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 29:

  # x1 = n1 + # x0, # x0 = n0 + # x, # x1 = n2 + # x, x1(n1 + n0 + n) = x0(n0 + n), x1(n1 + n0 + n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < # x1, n < # x1, n < # x0, # x + 1 < # x1, # x < # x1, # x < # x0, # x0 < # x1, n < # x, 2 + n < # x1, ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3), ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

Rule argument

Additional information

Proof step 29 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  # x1 = n2 + # x, # x0 = n0 + # x, # x1 = n1 + # x0, x1(n1 + n0 + n) = x0(n0 + n), x0(n0 + n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, 2 + n < # x1, n < # x, # x0 < # x1, # x < # x0, # x < # x1, # x + 1 < # x1, n < # x0, n < # x1, n + 1 < # x1, n + 1 < # x0, ∀ n3. n3 < # x → x(n3) = x0(n0 + n3), ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3) ⊦

Rule argument

Additional information

Proof step 30 from proof trans-List (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 31:

  # x0 = n0 + # x, n1 + # x0 = n2 + # x, x1(n1 + n0 + n) = x0(n0 + n), x0(n0 + n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 0, n2 ≠ 1, n + 1 < # x0, n + 1 < n1 + # x0, n < n1 + # x0, n < # x0, # x + 1 < n1 + # x0, # x < n1 + # x0, # x < # x0, # x0 < n1 + # x0, n < # x, 2 + n < n1 + # x0, ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3), ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

Rule argument

Additional information

Proof step 31 from proof trans-List (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  # x0 = n0 + # x, n1 + # x0 = n2 + # x, x1(n1 + n0 + n) = x0(n0 + n), x0(n0 + n) ≠ x1(n2 + n), n0 ≠ 0, n1 ≠ 0, n2 ≠ 1, n2 ≠ 0, n + 1 < # x0, n < n2 + # x, n < # x0, # x < # x0, n < # x, 2 + n < n2 + # x, n + 1 < n2 + # x, ∀ n3. n3 < # x0 → x0(n3) = x1(n1 + n3), ∀ n3. n3 < # x → x(n3) = x0(n0 + n3) ⊦

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) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m + n < m0 → m + n < m0 + n0; (s)

Proof step 32 from proof trans-List (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

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