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Proof step 5 from proof in-valid-eraseqp-valid-lref (previous step)

eqent.pnum < # apebs, valid-eraseq(apebs, q0), valid-peb(v, m, apebs[n]), valid-peb(apebs[eqent.pnum]) → eqent.lebref = apebs[eqent.pnum].vidhdr.vol × apebs[eqent.pnum].vidhdr.leb, ∀ apebs, v, m, n, lref. valid-eraseq(apebs, q0) ∧ valid-peb(v, m, apebs[n]) ∧ eq-entry(n, lref) ∈ q0 → lref = v × m, eq-entry(n, lref) = eqent ∨ eq-entry(n, lref) ∈ q0 ⊦ lref = v × m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
eqent = eq-entry(m0, lref0), m0 < # apebs, valid-eraseq(apebs, q0), valid-peb(v, m, apebs[n]), valid-peb(apebs[m0]) → lref0 = apebs[m0].vidhdr.vol × apebs[m0].vidhdr.leb, ∀ apebs, v, m, n, lref. valid-eraseq(apebs, q0) ∧ valid-peb(v, m, apebs[n]) ∧ eq-entry(n, lref) ∈ q0 → lref = v × m, eq-entry(n, lref) = eqent ∨ eq-entry(n, lref) ∈ q0 ⊦ lref = v × m

Rule argument

spec: erasequeueentry name: elim seq: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref) subst: [eqent, m, lref] → [eqent, m0, lref0]

Additional information

Used simplifier rules: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref); (elim) Used terms: eqent, m0, lref0