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Proof step 3 from proof lebrefs-append-none (previous step)

∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ (q0 + eq-entry(n, none) ').lebrefs ∪ (eqent.lebref ≠ none ⊃ ∅ ++ eqent.lebref;∅) = q0.lebrefs ∪ (eqent.lebref ≠ none ⊃ ∅ ++ eqent.lebref;∅)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
eqent = eq-entry(m, lref), ∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ (q0 + eq-entry(n, none) ').lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) = q0.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ 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, lref]

Additional information

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