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Proof step 10 from proof sorted-all (previous step)

n < # zbrar, n + m3 + 1 < # zbrar, (n + 1) + m3 < # zbrar ∧ (∀ m, m0. m < m0 ∧ ¬ m3 < m0 → zbrar[(n + 1) + m].key < zbrar[(n + 1) + m0].key) → sorted(zbrar, n + 1, m3 + 1), ∀ m, m0. m < m0 ∧ ¬ m3 + 1 < m0 → zbrar[n + m].key < zbrar[n + m0].key ⊦ sorted(zbrar, n + 1, m3 + 1) ∧ zbrar[n].key < zbrar[n + 1].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n + m3 + 1 < # zbrar, n < # zbrar, (m < m0 ∧ ¬ m3 < m0 → zbrar[(n + 1) + m].key < zbrar[(n + 1) + m0].key) → sorted(zbrar, n + 1, m3 + 1), ∀ m, m0. m < m0 ∧ ¬ m3 + 1 < m0 → zbrar[n + m].key < zbrar[n + m0].key ⊦ sorted(zbrar, n + 1, m3 + 1) ∧ zbrar[n].key < zbrar[n + 1].key

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