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Proof step 14 from proof in-concat (previous step)

m = n ∧ eq-entry(n, lref) ∈ q0 ∨ m0 = n ∧ eq-entry(n, lref0) ∈ q1 ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1)

Heuristic quantifier applied rule exists right

And obtained the following new goals

Proof step 15:
m = n ∧ eq-entry(n, lref) ∈ q0 ∨ m0 = n ∧ eq-entry(n, lref0) ∈ q1 ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q0 ∨ eqent ∈ q1), eq-entry(n, lref).pnum = n ∧ (eq-entry(n, lref) ∈ q0 ∨ eq-entry(n, lref) ∈ q1)

Rule argument

pos: right 1 terms: eq-entry(n, lref)

Additional information

Used terms: eq-entry(n, lref)