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Proof step 25 from proof valid-sqns-extract2-sqn-h (previous step)

apebs[n1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n1 < # apebs, n0 < # apebs, m1 < # apebs, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n1 ≤ m1, n0 ≤ m1, n1 < m1 ∧ n0 < m1 → valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn ++ avhdr0.sqn ⊦ n0 = n1, n1 < m1, ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ avhdr.sqn ++ avhdr0.sqn

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), apebs[m1] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), n0 < m1, n0 < # apebs, m1 < # apebs, n0 + 1 < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ⊦ n0 = m1, ∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn

Rule argument

Additional information

Used simplifier rules: ⊦ bag ++ n = bag2 ++ n ↔ bag = bag2; (s)%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]% ⊦ bag ∪ {n} = bag ++ n; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ m; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)