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Proof step 2 from proof in-concat (previous step)

⊦ (∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1)) ↔ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q0) ∨ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q1)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1) ⊦ ∃ eqent. eqent.pnum = n ∧ eqent ∈ q0, ∃ eqent. eqent.pnum = n ∧ eqent ∈ q1
Proof step 9:
(∃ eqent. eqent.pnum = n ∧ eqent ∈ q0) ∨ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q1) ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1)

Rule argument

pos: right 1

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