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Proof step 16 from proof in-concat (previous step)

eq-entry(n, lref0) ∈ q1 ⊦ eq-entry(n, lref) ∈ q1, eq-entry(n, lref) ∈ q0, ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: eq-entry(n, lref0)

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref).pnum = m; (s) Used terms: eq-entry(n, lref0)