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Proof step 12 from proof valid-global-sqn-invalidate (previous step)

m < # apebs, n ≤ valid-sqns(v, m0, apebs).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) = ∅, valid-peb(v, m0, apebs[m]), valid-peb(apeb)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: v, m0

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: v, m0