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Proof step 10 from proof in-append-other (previous step)

eqent = eq-entry(m, lref0), eqent ∈ q ⊦ m = n0, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 ∈ q ∨ eqent0 = eq-entry(n0, lref))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information