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Proof step 4 from proof mod-other-aux (previous step)

f = f(r; f(r)), (a[P] * R)[rs, f, g] ⊦ g(r) = a, (a[P] * true)[rs, f, g(r; a0)]

User applied rule insert equation

And obtained the following new goals

Proof step 5:
f = f(r; f(r)), (a[P] * R)[rs, f, g] ⊦ g(r) = a, (a[P] * true)[rs, f(r; f(r)), g(r; a0)]

Rule argument

pos: left 1 flags: [[2,2,3]]

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