 Solution

Let $R=\mathrm{diag}(1,e^{i\pi /12})$. Since the Hadamard gate maps the standard basis to $|+⟩,|-⟩$, the given gate is $Q=HRH$. Because $H^2=I$,Next, use $HZH=X$ on the target wire of the Controlled-Z gate:The control-zero block remains the identity, and the control-one block becomes $HZH=X$, proving the identity with the stated control convention. We have therefore constructed each gate in the assumed universal quantum gate set $\{\mathrm{CNOT},H,T\}$ using only $CZ,Q,H$.

Replacing every gate of a circuit by its displayed finite construction reproduces that circuit exactly. Thus $\{CZ,Q,H\}$ is universal under the usual arbitrary-accuracy convention: it can perform every approximation obtainable from the given universal set. The claim concerns approximation of arbitrary unitary operators; it does not assert that finitely many circuits exactly represent an uncountable continuum of gates. This is universality transfer by exact gate synthesis.