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Proof step 9 from proof valid-sqns-extract-sqn-h (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 10:
apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn, m1 < # apebs, n < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n ≤ m1 ⊦ ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ avhdr.sqn

Rule argument

pos: left 6

Additional information

Used simplifier rules: ⊦ 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)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ bag ∪ {n} = bag ++ n; (s)%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]% ⊦ bag ++ n = bag2 ++ n ↔ bag = bag2; (s)