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Proof step 8 from proof in-append-other (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information