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Proof step 6 from proof in-concat (previous step)

eq-entry(m, lref) ∈ q0 ∨ eq-entry(m, lref) ∈ q1 ⊦ ∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q1, ∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0

Heuristic quantifier applied rule exists right

And obtained the following new goals

Proof step 7:
eq-entry(m, lref) ∈ q0 ∨ eq-entry(m, lref) ∈ q1 ⊦ ∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0, ∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q1, eq-entry(m, lref).pnum = m ∧ eq-entry(m, lref) ∈ q0

Rule argument

pos: right 2 terms: eq-entry(m, lref)

Additional information

Used terms: eq-entry(m, lref)