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Proof step 5 from proof subseteq-lebrefs-subseteq (previous step)

eqent ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ q.lebrefs ∪ (eqent.lebref ≠ none ⊃ ∅ ++ eqent.lebref;∅) ⊆ q1.lebrefs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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