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

apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), ∀ 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) ⊦ valid-eraseq(q0), valid-eraseq(eq-entry(m0, lref) ' + q0), m0 ∈ q0, ∃ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: apebs, vtbl

Additional information

Used terms: apebs, vtbl