= Solution
The <controlled-NOT gate> is its own inverse, so $\mathrm{CNOT}^2=I$. Every remaining nontrivial gate acts on one wire. By the <tensor product> rule, any circuit on a product input stays a <product state>; in particular, starting at $|00\rangle$ it cannot produce a <Bell state>.
This also prevents arbitrary approximation, not merely exact preparation. For normalized single-qubit <vectors> $|u\rangle,|v\rangle$,
$$
|\langle\Phi^+|u\otimes v\rangle|^2=\tfrac12|u_0v_0+u_1v_1|^2\leq\tfrac12
$$
by the <Cauchy-Schwarz inequality>. Hence the <quantum fidelity> with $|\Phi^+\rangle$ is bounded away from one for every reachable state. \b[This gate set is not universal.] The obstruction is preservation of <product states>, a special case of the existing <product-state obstruction to SWAP universality>.
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