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Proof step 13 from proof in-append (
previous step
)
n = n0 ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 14
:
⊦ ∃ eqent. eqent.pnum = n0 ∧ (eqent ∈ q ∨ eqent = eq-entry(eqent.pnum, lref))
Rule argument
Additional information