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Proof step 2 from proof valid-sqns-extract-sqn (previous step)

n < # apebs, valid-peb(v, m, apebs[n]) ⊦ ∃ bag. valid-sqns(v, m, apebs) = bag ++ apebs[n].vidhdr.sqn

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # apebs, valid-peb(v, m, apebs[n]) ⊦ ∃ bag. valid-sqns(v, m, apebs, # apebs) = bag ++ apebs[n].vidhdr.sqn

Rule argument

Additional information

Used simplifier rules: valid-sqns: ⊦ valid-sqns(v, m, apebs) = valid-sqns(v, m, apebs, # apebs);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic