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

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref)) ⊦ n = n0, ∃ eqent. eqent.pnum = n ∧ eqent ∈ q
Proof step 7:
(∃ eqent. eqent.pnum = n ∧ eqent ∈ q) ∨ n = n0 ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref))

Rule argument

pos: right 1

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