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Proof step 10 from proof write-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], a[r => o0][rs1, f, g] ⊦ a[r => o1][rs1, f(r; o1), g], (a[r => o1] * R)[rs1 ∪ rs0, f(r; o1), g] ∧ f(r) = o0

User applied rule case distinction

And obtained the following new goals

Proof step 11:
rs1 ∩ rs0 = ∅, a[r => o0][rs1, f, g], R[rs0, f, g] ⊦ a[r => o1][rs1, f(r; o1), g], (a[r => o1] * R)[rs1 ∪ rs0, f(r; o1), g]
Proof step 37:
rs1 ∩ rs0 = ∅, a[r => o0][rs1, f, g], R[rs0, f, g] ⊦ f(r) = o0, a[r => o1][rs1, f(r; o1), g]

Rule argument

pos: right 2

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