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Proof step 2 from proof in-append-other (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref))) ↔ (∃ 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 ' ↔ eqent = eqent0; (s)% ⊦ eqent ∈ q + q0 ↔ eqent ∈ q ∨ eqent ∈ q0; (s)