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Proof step 19 from proof indices-erasing-eraseq (previous step)

wlar1[eqent.pnum] = wl-entry(m1, stat), wlar0[eqent.pnum] = wl-entry(m0, erasing), eqent.pnum < # wlar0, indices-erasing(wlar1, nx0), indices-erasing(wlar0, nx0), valid-eraseq(wlar1, q1), eqent ∈ q0, ∀ eqent. eqent ∈ q0 → eqent ∈ q1 ⊦ eqent.pnum < # wlar1 ∧ stat = erasing

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
wlar0[m] = wl-entry(m0, erasing), wlar1[m] = wl-entry(m1, stat), eqent = eq-entry(m, lref), m < # wlar0, indices-erasing(wlar0, nx0), indices-erasing(wlar1, nx0), valid-eraseq(wlar1, q1), eqent ∈ q0, ∀ eqent. eqent ∈ q0 → eqent ∈ q1 ⊦ m < # wlar1 ∧ stat = erasing

Rule argument

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

Additional information

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