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Proof step 10 from proof valid-global-sqn-invalidate (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: v, m0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # apebs ⊦ apebs[n] = apeb ∧ valid-peb(v, m, apeb) → valid-sqns(v, m, apebs) ≠ ∅; (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)% ⊦ bag -- n ≠ ∅ → ¬ bag.max < (bag -- n).max; (s) Used terms: v, m0