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Proof step 5 from proof valid-peb-in-sqnums-h (previous step)

apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n0 < # apebs, n < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n0 ≤ n, ∀ apebs, n0, aehdr, avhdr, buf, m0, boolvar, v, m. apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar) ∧ n0 < n ∧ n0 < # apebs ∧ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ n ≤ # apebs → avhdr.sqn ∈ valid-sqns(v, m, apebs, n) ⊦ avhdr.sqn ∈ valid-sqns(v, m, apebs, n), avhdr.sqn ∈ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: apebs, n0, aehdr, avhdr, buf, m0, boolvar, v, m abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ n ∈ {n0} ↔ n = n0; (s) Used terms: apebs, n0, aehdr, avhdr, buf, m0, boolvar, v, m