Proof step 1 from proof allgerange-recup (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(all≥(x, m, n, a) ↔ ¬ x[n - 1] < a ∧ all≥(x, m, n - 1, a)), n < # x, m < n ⊦

Rule argument

Additional information

Proof step 2 from proof allgerange-recup (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  x ≠ [], m < # x, n < # x, m < n, m + 1 < # x ⊦ (∀ n0. m ≤ n0 ∧ n0 < n ∧ n0 < # x → ¬ x[n0] < a) ↔ ¬ x[n - 1] < a ∧ (∀ n0. m ≤ n0 ∧ n0 + 1 < n ∧ n0 < # x → ¬ x[n0] < a)

Rule argument

Additional information

Used simplifier rules: allge-range-def: ⊦ all≥(x, m, n, a) ↔ (∀ n0. m ≤ n0 ∧ n0 < n ∧ n0 < # x → ¬ x[n0] < a); ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) ⊦ # x = 0 ↔ x = []; (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 f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 3 from proof allgerange-recup (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 4:

  n = m0 + 1, n ≠ 0, x ≠ [], m + 1 < # x, m < n, n < # x, m < # x ⊦ (∀ n0. m ≤ n0 ∧ n0 < n ∧ n0 < # x → ¬ x[n0] < a) ↔ ¬ x[m0] < a ∧ (∀ n0. m ≤ n0 ∧ n0 + 1 < n ∧ n0 < # x → ¬ x[n0] < a)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ # x = 0 ↔ x = []; (s) ⊦ ¬ n < 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0

Proof step 4 from proof allgerange-recup (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  x ≠ [], m + 1 < # x, m < # x, m0 + 1 < # x, m ≤ m0 ⊦ (∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a) ↔ ¬ x[m0] < a ∧ (∀ n0. m ≤ n0 ∧ n0 < m0 ∧ n0 < # x → ¬ x[n0] < a)

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) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m + 1 ≠ 0; (s)

Proof step 5 from proof allgerange-recup (previous step) (back to tree)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

• Proof step 6:

  x ≠ [], m0 + 1 < # x, m < # x, m + 1 < # x, m ≤ m0, ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a ⊦ ¬ x[m0] < a ∧ (∀ n0. m ≤ n0 ∧ n0 < m0 ∧ n0 < # x → ¬ x[n0] < a)

• Proof step 11:

  x ≠ [], m0 + 1 < # x, m < # x, m + 1 < # x, m ≤ m0, ¬ x[m0] < a, ∀ n0. m ≤ n0 ∧ n0 < m0 ∧ n0 < # x → ¬ x[n0] < a ⊦ ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a

Rule argument

Additional information

Proof step 6 from proof allgerange-recup (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  x ≠ [], m + 1 < # x, m < # x, m0 + 1 < # x, m ≤ m0, ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a ⊦ ¬ x[m0] < a ∧ (m ≤ n0 ∧ n0 < m0 ∧ n0 < # x → ¬ x[n0] < a)

Rule argument

Additional information

Proof step 7 from proof allgerange-recup (previous step) (back to tree)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

• Proof step 8:

  x ≠ [], x[m0] < a, m0 + 1 < # x, m < # x, m + 1 < # x, m ≤ m0, ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a ⊦

• Proof step 9:

  x ≠ [], x[n0] < a, n0 < m0, n0 < # x, m0 + 1 < # x, m < # x, m + 1 < # x, m ≤ n0, m ≤ m0, ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a ⊦

Rule argument

Additional information

Proof step 8 from proof allgerange-recup (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) ⊦ ¬ m < n → ¬ m + m0 < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0

Proof step 9 from proof allgerange-recup (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  x ≠ [], m < m0, x[n0] < a, n0 < m0, n0 < # x, m0 + 1 < # x, m < # x, m + 1 < # x, m ≤ n0, ∀ n0. m ≤ n0 ∧ ¬ m0 < n0 ∧ n0 < # x → ¬ x[n0] < a ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ m < n → n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 10 from proof allgerange-recup (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s) Used terms: n0

Proof step 11 from proof allgerange-recup (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  x ≠ [], x[n0] < a, n0 < # x, m + 1 < # x, m < # x, m0 + 1 < # x, m ≤ n0, n0 ≤ m0, m ≤ m0, ¬ x[m0] < a, ∀ n0. m ≤ n0 ∧ n0 < m0 ∧ n0 < # x → ¬ x[n0] < a ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 12 from proof allgerange-recup (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used terms: n0