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

n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, 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) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

User applied rule case distinction

And obtained the following new goals

Proof step 10:
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}

Rule argument

pos: right 1

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]