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Proof step 1 from proof aux-known (conclusion)

⊦ (a0[r => o] * R)[rs, f, g] ↔ (a1[r => o] * R)[rs, f, g(r; a1)] ∧ g(r) = a0

User applied rule case distinction

And obtained the following new goals

Proof step 2:
(a0[r => o] * R)[rs, f, g] ⊦ (a1[r => o] * R)[rs, f, g(r; a1)] ∧ g(r) = a0
Proof step 46:
g(r) = a0, (a1[r => o] * R)[rs, f, g(r; a1)] ⊦ (a0[r => o] * R)[rs, f, g]

Rule argument

pos: right 1

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