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Proof step 14 from proof valid-eraseq-strong-imp-p (previous step)

apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq(apebs, q0), valid-eraseq-strong(apebs, vtbl, q0), valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)), ∀ v, m. (valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → avhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ lref = avhdr.vol × avhdr.leb, m0 ∈ q0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: avhdr.vol, avhdr.leb

Additional information

Used simplifier rules: ⊦ valid-peb(apeb) → (valid-peb(v, m, apeb) ↔ v = apeb.vidhdr.vol ∧ m = apeb.vidhdr.leb); (ss)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: avhdr.vol, avhdr.leb