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Proof step 19 from proof valid-sqns-subseteq-h (previous step)

apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 + 1 < # apebs, n0 < n, n < # apebs, n0 < # apebs, count(n1, valid-sqns(v, m, apebs, n) ++ avhdr.sqn) < count(n1, valid-sqns(v, m, apebs, n0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n0 ≤ n + 1, count(n1, valid-sqns(v, m, apebs, n0)) ≤ count(n1, valid-sqns(v, m, apebs, n)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), count(n1, valid-sqns(v, m, apebs, n) ++ avhdr.sqn) < count(n1, valid-sqns(v, m, apebs, n0)), n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, count(n1, valid-sqns(v, m, apebs, n) ++ avhdr.sqn) < count(n1, valid-sqns(v, m, apebs, n)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), count(n1, valid-sqns(v, m, apebs, n0)) ≤ count(n1, valid-sqns(v, m, apebs, n)), n0 ≤ n + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat