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

apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 < # apebs, valid-peb(v, m, apebs[n0]), ∀ apebs, v, m. n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) = ∅ ∨ (∃ n1. n1 < n0 ∧ valid-peb(v, m, apebs[n1]) ∧ apebs[n1].vidhdr.sqn = valid-sqns(v, m, apebs, n0).max) ⊦ ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = (valid-sqns(v, m, apebs, n0) ++ avhdr.sqn).max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), ∀ apebs, v, m. n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) = ∅ ∨ (∃ n1. n1 < n0 ∧ valid-peb(v, m, apebs[n1]) ∧ apebs[n1].vidhdr.sqn = valid-sqns(v, m, apebs, n0).max) ⊦ ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = (valid-sqns(v, m, apebs, n0) ++ avhdr.sqn).max

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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