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Proof step 9 from proof sorted-all (previous step)

n + m3 + 1 < # zbrar, n < # zbrar, ∀ n0, zbrar0. n0 + m3 < # zbrar0 ∧ (∀ m, m0. m < m0 ∧ ¬ m3 < m0 → zbrar0[n0 + m].key < zbrar0[n0 + m0].key) → sorted(zbrar0, n0, 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

User applied rule all left

And obtained the following new goals

Proof step 10:
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

Rule argument

pos: left 3 terms: n + 1, zbrar abort: F discard: T computed: T

Additional information

Used terms: n + 1, zbrar