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Proof step 5 from proof valid-eraseqd-nodups-middle2 (previous step)

valid-eraseq(q + eq-entry(m, lref) ' + q1), m ∈ (eqent ' + q) + q1, ∀ m0, lref0, q0. ¬ valid-eraseq(q + eq-entry(m0, lref0) ' + q0) ∨ ¬ m0 ∈ q + q0 ⊦ eqent.pnum = m, 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, lref1), valid-eraseq(q + eq-entry(m, lref) ' + q1), m ∈ (eqent ' + q) + q1, ∀ m0, lref0, q0. ¬ valid-eraseq(q + eq-entry(m0, lref0) ' + q0) ∨ ¬ m0 ∈ q + q0 ⊦ m1 = m, 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, lref1]

Additional information

Used simplifier rules: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref); (elim) Used terms: eqent, m1, lref1