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Proof step 14 from proof sorted-lt-h (previous step)

m0 = 0, n < # zbrar, zbrar[n].key < zbrar[n + 1].key, sorted(zbrar, n + 1, m + 1), m4 ≤ m, 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 + m4 + 1].key ↔ ¬ m4 < m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
zbrar[n].key < zbrar[n + 1].key, n < # zbrar, zbrar[n + m4 + 1].key < zbrar[n + 1].key, sorted(zbrar, n + 1, m + 1), m4 ≤ m, ∀ 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].key < zbrar[n + m4 + 1].key

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ key0 < key ∧ key0 < key1 → key < key1; (forward) from specification oelem[key-order]% ⊦ 0 ≤ m; (s)% ⊦ ¬ m < 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)