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Proof step 9 from proof lift-ex-o (previous step)

⊦ a[_∃ oP](mk(rs, f, g)) ↔ (∃ o. a[oP(o)](mk(rs, f, g)))

User applied rule case distinction

And obtained the following new goals

Proof step 10:
a[_∃ oP](mk(rs, f, g)) ⊦ ∃ o. a[oP(o)](mk(rs, f, g))
Proof step 16:
∃ o. a[oP(o)](mk(rs, f, g)) ⊦ a[_∃ oP](mk(rs, f, g))

Rule argument

pos: right 1

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