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Proof step 20 from proof valid-eraseq-stronger-one (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m < # apebs, ∀ v, m1. (valid-peb(v, m1, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ↔ v0 × m2 = v × m1) ∧ (v0 × m2 = v × m1 → avhdr.sqn ≠ valid-sqns(v, m1, apebs).max) ⊦ valid-peb(v0, m2, mkpeb(aehdr, avhdr, buf, m0, boolvar))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: v0, m2

Additional information

Used terms: v0, m2