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Proof step 12 from proof lift-emp (previous step)

⊦ mk(rs, f) = ∅ ∧ (∀ r. r ∈ rs → g(r) = a) ↔ mk(rs, f, g) = ∅

User applied rule case distinction

And obtained the following new goals

Proof step 13:
mk(rs, f) = ∅, ∀ r. r ∈ rs → g(r) = a ⊦ mk(rs, f, g) = ∅
Proof step 21:
mk(rs, f, g) = ∅ ⊦ mk(rs, f) = ∅ ∧ (∀ r. r ∈ rs → g(r) = a)

Rule argument

pos: right 1

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