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Proof step 4 from proof valid-global-sqn-add-valid (previous step)

apeb = mkpeb(aehdr, avhdr, buf, m1, boolvar), avhdr.sqn < valid-sqns(v, m0, apebs[m, apeb]).max, avhdr.sqn < n, m < # apebs, valid-peb(apeb), n ≤ valid-sqns(v, m0, apebs[m, apeb]).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs[m, apeb]) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
avhdr.sqn < valid-sqns(v, m0, apebs[m, mkpeb(aehdr, avhdr, buf, m1, boolvar)]).max, avhdr.sqn < n, m < # apebs, valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)), n ≤ valid-sqns(v, m0, apebs[m, mkpeb(aehdr, avhdr, buf, m1, boolvar)]).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs[m, mkpeb(aehdr, avhdr, buf, m1, boolvar)]) = ∅

Rule argument

Additional information

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