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Proof step 6 from proof sorted-lt-h (previous step)

m2 = m +1, sorted(zbrar, n, m2 + 1), m1 ≤ m2, m0 ≤ m2, ∀ m0, m1, zbrar, n. m0 < m2 ∧ m1 < m2 ∧ sorted(zbrar, n, m2) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1) ⊦ zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
zbrar[n].key < zbrar[n + 1].key, n < # zbrar, sorted(zbrar, n + 1, m + 1), m1 ≤ m + 1, m0 ≤ m + 1, ∀ m2, m3, zbrar0, n0. ¬ m < m2 ∧ ¬ m < m3 ∧ sorted(zbrar0, n0, m + 1) → (zbrar0[n0 + m2].key < zbrar0[n0 + m3].key ↔ m2 < m3) ⊦ zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1

Rule argument

Additional information

Used simplifier rules: sorted-rec: ⊦ sorted(zbrar, n, m + 2) ↔ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m + 1);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%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)