Evaluate $f = x \to (y \equiv z)$ only via $\oplus$
I have to find an equivalent expression using only $\oplus$
$y \equiv z = y \oplus z \oplus 1$
$x \to y = 1 \oplus x \oplus xy$
$f = x \to (y \equiv z) = 1 \oplus x \oplus x (y \oplus z \oplus 1) = 1 \oplus x \oplus xy \oplus xz \oplus x = 1 \oplus xy \oplus xz = 1 \oplus x(y \oplus z)$
But I can not get rid of the conjunction $x(y \oplus z)$
How can one get rid of the conjunction?
Any ideas?
Like your previous question, this is impossible.
Since the $\oplus$ is associative, any expression built up using $\oplus$ as the only operator can be treated as one big generalized operator that works on a collection of terms, where all terms are propositional variables.
Moreover, it is easy to show that this generalized $\oplus$ is true if and only if an odd number of its terms are true. So, for any variable that occurs an even number of times, the value of the expression will not depend on the value of that variable. And for any variable that occurs an odd number of times, the value of the expression will be different for different values of that variable.
Example:
Say we have an expression like $$((x \oplus y) \oplus (x \oplus z)) \oplus y$$
First of all, since $\oplus$ is associative, it really doesn;t matter in which we work out the values, and so we can treat this simply as:
$$x \oplus y \oplus x \oplus z \oplus y$$
Second, we see that $x$ occurs an even number of times, and so the value of the expression does not change depending on the value of $x$. Same goes for $y$. But $z$ occurs an odd number of times, and so the expression is fully dependent on $z$ in the sense that if $z$ is true, the expression takes on a different truth-value then when $z$ is false. We can easily verify these claims:
$$x \oplus y \oplus x \oplus z \oplus y = x \oplus x \oplus y \oplus y \oplus z = 0 \oplus 0 \oplus z = 0 \oplus z = z$$
Of course, that is just an example, but by induction you can prove what I claimed above in general.
Now, let's look at your claim:
$$x \rightarrow (y \leftrightarrow z)$$
The value of this expression clearly depends on $x$, but does not fully depend on $x$, for $x$ is false, the expression is true, but if $x$ is true, then the expression's value depends on the values of $y$ and $z$.
So, you cannot capture that kind of function using $\oplus$ alone.