= Solution
Interpret $h_A,h_B$ as <cause-specific hazards>: their conditioning population consists of individuals still free of both events. Write
$$
H_A(t)=\int_0^t h_A(u)\,du,\qquad H_B(t)=\int_0^t h_B(u)\,du.
$$
In a short interval, either event removes an individual from that population. Hence the <survivor function> satisfies $S'(t)=-S(t)[h_A(t)+h_B(t)]$, with $S(0)=1$. Integrating gives
$$
\boxed{\Pr(T>t)=S(t)=\exp\{-H_A(t)-H_B(t)\}.}
$$
No independence assumption on hypothetical latent event times is needed: this follows directly from the specified <cause-specific hazards>.
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