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Proof step 11 from proof valid-eraseq-stronger-imp-strong (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: v, m

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s) Used terms: v, m