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

n0 + 1 < # apebs, n0 < n, n < # apebs, n0 < # apebs, valid-peb(v, m, apebs[n]), n0 ≤ n + 1, valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ {apebs[n].vidhdr.sqn}

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, valid-peb(v, m, apebs[n]), n0 ≤ n + 1, valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ++ apebs[n].vidhdr.sqn

Rule argument

Additional information

Used simplifier rules: 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)