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Proof step 1 from proof lift-unknown (conclusion)

⊦ (r ==>_ * R)[rs, f, g] ↔ (∃ a. (a[r =>_] * R)[rs, f, g])

User applied rule case distinction

And obtained the following new goals

Proof step 2:
(r ==>_ * R)[rs, f, g] ⊦ ∃ a. (a[r =>_] * R)[rs, f, g]
Proof step 41:
∃ a. (a[r =>_] * R)[rs, f, g] ⊦ (r ==>_ * R)[rs, f, g]

Rule argument

pos: right 1

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