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Proof step 5 from proof valid-sqns-nonempty-extract-max (previous step)

# apebs ≤ # apebs ∧ valid-sqns(v, m, apebs, # apebs) ≠ ∅ → (∃ n0. n0 < # apebs ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max) ⊦ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max, n < # apebs, valid-peb(v, m, apebs[n]) ⊦ 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: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ m; (s)