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Proof step 11 from proof sorted-all (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 12:
m < m0, n < # zbrar, n + m3 + 1 < # zbrar, ∀ m, m0. m < m0 ∧ ¬ m3 + 1 < m0 → zbrar[n + m].key < zbrar[n + m0].key ⊦ zbrar[(n + 1) + m].key < zbrar[(n + 1) + m0].key, m3 < m0, sorted(zbrar, n + 1, m3 + 1) ∧ zbrar[n].key < zbrar[n + 1].key
Proof step 14:
n < # zbrar, n + m3 + 1 < # zbrar, 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

Proof step 11 from proof sorted-all (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information