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Proof step 10 from proof valid-sqns-extract-sqn-h (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 11:
apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), valid-sqns(v, m, apebs, m1) = bag ++ avhdr.sqn, n < # apebs, m1 < # apebs, valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, apebs[m1]), n ≤ m1 ⊦ ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ avhdr.sqn

Rule argument

fma: valid-peb(v, m, apebs[m1])

Additional information

Used simplifier rules: ⊦ ∅ ∪ bag0 = bag0; (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) Used formulas: valid-peb(v, m, apebs[m1])