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Proof step 5 from proof valid-eraseq-strong-imp-p (previous step)

eqent.pnum < # apebs, valid-eraseq-strong(apebs, vtbl, q0), ∀ v, m. (valid-peb(v, m, apebs[eqent.pnum]) ↔ eqent.lebref = v × m) ∧ (eqent.lebref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[eqent.pnum].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ∀ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0) → valid-eraseq(apebs, q0) ⊦ valid-eraseq(apebs, eqent ' + q0), eqent.pnum ∈ q0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
eqent = eq-entry(m0, lref), m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), ∀ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0) → valid-eraseq(apebs, q0), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ valid-eraseq(apebs, eqent ' + q0), m0 ∈ q0

Rule argument

spec: erasequeueentry name: elim seq: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref) subst: [eqent, m, lref] → [eqent, m0, lref]

Additional information

Used simplifier rules: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref); (elim) Used terms: eqent, m0, lref