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Proof step 23 from proof lift-ex-r (previous step)

rP(r)(mk(rs, f)), ∀ r. r ∈ rs → g(r) = a ⊦ (∃ r. rP(r)(mk(rs, f))) ∧ (r0 ∈ rs → g(r0) = a)

User applied rule case distinction

And obtained the following new goals

Proof step 24:
rP(r)(mk(rs, f)), ∀ r. r ∈ rs → g(r) = a ⊦ ∃ r. rP(r)(mk(rs, f))
Proof step 25:
rP(r)(mk(rs, f)), r0 ∈ rs, ∀ r. r ∈ rs → g(r) = a ⊦ g(r0) = a

Rule argument

pos: right 1

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