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Proof step 3 from proof in-prepend (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
eqent ∈ q ⊦ eqent.pnum = n0, eqent = eq-entry(n0, lref), ∃ eqent0. eqent0.pnum = eqent.pnum ∧ eqent0 ∈ q

Rule argument

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref).pnum = m; (s)