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Proof step 9 from proof in-append-other (previous step)

eqent ∈ q ⊦ eqent.pnum = n0, ∃ eqent0. eqent0.pnum = eqent.pnum ∧ (eqent0 ∈ q ∨ eqent0 = eq-entry(n0, lref))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
eqent = eq-entry(m, lref0), eqent ∈ q ⊦ m = n0, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 ∈ q ∨ eqent0 = eq-entry(n0, lref))

Rule argument

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

Additional information

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