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

apebs[m1] = mkpeb(aehdr, avhdr, buf, m2, boolvar), m1 < # apebs, valid-eraseq-strong(apebs, to-vtbl(vols), q0), ∀ v0, m0. (valid-peb(v0, m0, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ↔ v × m = v0 × m0) ∧ (v × m = v0 × m0 ∧ v0 ∈ vols ∧ m0 < # vols[v0] → avhdr.sqn ≠ valid-sqns(v0, m0, apebs).max), ∀ apebs, vols, n, v, m. valid-eraseq-strong(apebs, to-vtbl(vols), q0) ∧ eq-entry(n, v × m) ∈ q0 → valid-peb(v, m, apebs[n]) ∧ (v ∈ vols ∧ m < # vols[v] → apebs[n].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ m1 ∈ q0, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ∧ (v ∈ vols ∧ m < # vols[v] → avhdr.sqn ≠ valid-sqns(v, m, apebs).max)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 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