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The unique fixed point principle says that a suitably guarded recursion has a unique solution in the chosen CSP semantic model. In particular, for a guarded process expression $F(X)$,The same principle applies to a mutually recursive system: two tuples satisfying the same guarded equations are componentwise equal. Thus an algebraic proof may show that a candidate satisfies the defining equations, instead of comparing arbitrarily long executions individually.

A guarded recursion must make visible progress before re-entering its recursive dependency. For an equation built from CSP prefixing and CSP choice, the usual sufficient condition is that every recursive occurrence lies behind a visible CSP prefixing operation. For mutually recursive equations, every dependency cycle must pass such a guard. The guard must survive enclosing operators: an event subsequently removed by CSP hiding does not provide visible progress. Equivalently, in the usual semantic formulation the recursion functional must be contractive in observable depth. Merely seeing an arrow somewhere before the variable is insufficient.

Let $X=P{}_{\{a,b,d\}}{\Vert }_{\{a,b,c\}}Q$. In this CSP alphabetized parallel composition, $a,b$ require both participants, $d$ belongs only to the left participant and $c$ only to the right participant. Expand the reachable states using CSP prefixing and CSP external choice:For example, after the synchronized $b$, the left participant is back at $P$ but the right participant is at $c\to Q$. The left participant's proposed $a$ is blocked until the right participant has performed $c$. Similarly, after $a,d$, only the synchronized $b$ is possible. Thus there is no extra interleaving to include.

Eliminating $Y,W,Z$ gives the guarded equationConsequently choose the sequential CSP processThe unique fixed point principle first establishes $R=X$.

Now let $E=X{\Theta }_{\{c\}}X$. The CSP exception operator follows its left operand until that operand performs $c$, includes this $c$ in the visible behaviour, and then switches to its right operand. Applying its CSP step law to the four-state expansion yieldsThe switch occurs exactly where the ordinary execution already returns to $X$. Therefore $E=X=R$. The answer uses neither a CSP exception operator nor a CSP parallel composition operator, and the argument gives equality in all three standard CSP semantic models.