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Proof step 10 from proof valid-sqns-nonempty-extract-max-h (previous step)

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, ¬ n0 < n0 ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = (valid-sqns(v, m, apebs, n0) ++ avhdr.sqn).max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), avhdr.sqn < valid-sqns(v, m, apebs, n0).max, 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) ⊦ valid-sqns(v, m, apebs, n0) = ∅, ∃ n. ¬ n0 < n ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, n0).max

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m > n ↔ n < m; (s)% ⊦ bag ≠ ∅ → (bag ++ m).max = (bag.max > m ⊃ bag.max;m); (s)% ⊦ m = (bag ++ m).max ↔ ¬(bag ≠ ∅ ∧ m < bag.max); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ ¬ n < n; (s)