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Proof step 17 from proof valid-global-sqn-next-max (previous step)

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 ⊦

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: v0, m5

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ (∅ ++ n).max = n; (s)% ⊦ ¬ n < n; (s) Used terms: v0, m5