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Proof step 1 from proof in-prepend (conclusion)

⊦ n ∈ eq-entry(n0, lref) ' + q ↔ n ∈ q ∨ n = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∃ eqent. eqent.pnum = n ∧ (eqent = eq-entry(n0, lref) ∨ eqent ∈ q)) ↔ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q) ∨ n = n0

Rule argument

Additional information

Used simplifier rules: queue-in: ⊦ n ∈ q ↔ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q);%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ eqent ∈ eqent0 ' + q ↔ eqent = eqent0 ∨ eqent ∈ q; (s)