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Proof step 1 from proof in-concat (conclusion)

⊦ n ∈ q0 + q1 ↔ n ∈ q0 ∨ n ∈ q1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1)) ↔ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q0) ∨ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q1)

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 ∈ q + q0 ↔ eqent ∈ q ∨ eqent ∈ q0; (s)