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Proof step 2 from proof write-known (previous step)

⊦ (a[r => o0] * R)[rs, f, g] ↔ (a[r => o1] * R)[rs, f(r; o1), g] ∧ f(r) = o0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
(a[r => o0] * R)[rs, f, g] ⊦ (a[r => o1] * R)[rs, f(r; o1), g] ∧ f(r) = o0
Proof step 77:
f(r) = o0, (a[r => o1] * R)[rs, f(r; o1), g] ⊦ (a[r => o0] * R)[rs, f, g]

Rule argument

pos: right 1

Additional information