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Proof step 3 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

# apebs = # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-recs(apebs, wlar, recs), valid-wl(apebs, wlar) ⊦ valid-sqns(v, m, apebs) = ∅, v × m ∈ recs

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ valid-sqns(v, m, apebs) ≠ ∅ → (∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)
Proof step 5:
# apebs = # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-recs(apebs, wlar, recs), valid-wl(apebs, wlar), valid-sqns(v, m, apebs) ≠ ∅ → (∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max) ⊦ valid-sqns(v, m, apebs) = ∅, v × m ∈ recs

Rule argument

spec: sqns name: valid-sqns-nonempty-extract-max seq: ⊦ valid-sqns(v, m, apebs) ≠ ∅ → (∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max) subst: [v, m, apebs] → [v, m, apebs]

Additional information

Used terms: v, m, apebs