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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: valid-sqns: ⊦ valid-sqns(v, m, apebs) = valid-sqns(v, m, apebs, # apebs);