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Proof step 4 from proof lebrefs-append-none (previous step)

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;∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ (q0 + eq-entry(n, none) ').lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) = q0.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅)

Proof step 4 from proof lebrefs-append-none (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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