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Proof step 3 from proof sorted-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n + m0 < # zbrar, ∀ m, m1. m < m1 ∧ ¬ m0 < m1 → zbrar[n + m].key < zbrar[n + m1].key, ∀ n, zbrar. n + m0 ≤ # zbrar ∧ (∀ m2, m1. m2 < m1 ∧ m1 < m0 → zbrar[n + m2].key < zbrar[n + m1].key) → sorted(zbrar, n, m0) ⊦ sorted(zbrar, n, m0 + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ 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% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)