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

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

User applied rule case distinction

And obtained the following new goals

Proof step 11:
m = n, eq-entry(n, lref0) ∈ q ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref))
Proof step 13:
n = n0 ⊦ ∃ eqent. eqent.pnum = n ∧ (eqent ∈ q ∨ eqent = eq-entry(n0, lref))

Rule argument

pos: left 1

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