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Proof step 6 from proof mod-other-aux (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 4:
f = f(r ; f(r)), (a[P] * R)[rs, f, g] ⊦ g(r) = a, (a[P] * true)[rs, f, g(r ; a0)]

Rule argument

[fma-arg: f = f(r ; f(r))]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (f = f(r ; f(r)))]]