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Proof step 16 from proof valid-global-sqn-next-max (previous step)

apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), m < # apebs, m1 < n, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), n ≤ (valid-sqns(v0, m5, apebs, m) ++ m1).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), m1 < (valid-sqns(v0, m5, apebs, m) ++ m1).max, m < # apebs, m1 < n, m1 < valid-sqns(v0, m5, apebs, m).max, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), n ≤ valid-sqns(v0, m5, apebs, m).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < (bag ++ m).max → (bag ++ m).max = bag.max; (s)% ⊦ 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