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Proof step 5 from proof valid-eraseqd-nodups1 (previous step)

valid-eraseq(q + q1), m ∈ q1, m ∈ eqent ' + q, ∀ q0, m0. ¬ valid-eraseq(q + q0) ∨ ¬ m0 ∈ q0 ∨ ¬ m0 ∈ q ⊦ eqent.pnum ∈ q, eqent.pnum ∈ q1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
eqent = eq-entry(m1, lref), valid-eraseq(q + q1), m ∈ eqent ' + q, m ∈ q1, ∀ q0, m0. ¬ valid-eraseq(q + q0) ∨ ¬ m0 ∈ q0 ∨ ¬ m0 ∈ q ⊦ m1 ∈ q1, m1 ∈ q

Rule argument

spec: erasequeueentry name: elim seq: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref) subst: [eqent, m, lref] → [eqent, m1, lref]

Additional information

Used simplifier rules: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref); (elim) Used terms: eqent, m1, lref