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Proof step 6 from proof valid-eraseqd-append (previous step)

∀ m, lref. valid-eraseq(q0 + eq-entry(m, lref) ') ↔ ¬ m ∈ q0 ∧ valid-eraseq(q0) ⊦ ¬ eqent.pnum ∈ q0 + eq-entry(m, lref) ' ∧ valid-eraseq(q0 + eq-entry(m, lref) ') ↔ ¬ m ∈ eqent ' + q0 ∧ ¬ eqent.pnum ∈ q0 ∧ valid-eraseq(q0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
eqent = eq-entry(m0, lref0), ∀ m, lref. valid-eraseq(q0 + eq-entry(m, lref) ') ↔ ¬ m ∈ q0 ∧ valid-eraseq(q0) ⊦ ¬ m0 ∈ q0 + eq-entry(m, lref) ' ∧ valid-eraseq(q0 + eq-entry(m, lref) ') ↔ ¬ m ∈ eqent ' + q0 ∧ ¬ m0 ∈ q0 ∧ valid-eraseq(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, lref0]

Additional information

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