Homomorphic & Functional encryption: Mapping unencrypted outputs to encrypted outputs using existing data
Actually, what you asked is not unusual, even more, it is always true.

Since $Eval(f, Enc(dA)) = Enc(f(dA))$ and $Eval(f, Enc(dU)) = Enc(f(dU))$, if $Eval(f, Enc(dA)) = Eval(f, Enc(dU))$, then $ Enc(f(dA)) =  Enc(f(dU))$.

Now, if you use the decryption algorithm in both sides, you get $f(dA) = f(dU)$.

The same argument applies if we use $Enc(f)$ instead of $f$.

The point now is whether this information is useful or not, because the probability that $Eval(f, Enc(dA)) = Eval(f, Enc(dU))$ is very small: even if $dA = dU$, you are supposed to have $Enc(dA) \not = Enc(dU)$, therefore, the evaluation algorithm is likely to give you different outputs.

So, you would probably never see this event and, consequently, you would not be able to use the conclusion (the right side of the implication).