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Proof step 8 from proof valid-sqns-nonempty-extract-max (previous step)

avhdr.sqn = valid-sqns(v, m, apebs, # apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊦ valid-sqns(v, m, apebs, # apebs) = ∅, ∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = avhdr.sqn

Heuristic quantifier applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: n