Renewal equation for joint forward and backward renewal processes Let $T = \{T_n; n=0,1,\ldots\}$ be a renewal sequence with inter-renewal distribuition $F.$ 
Let $$A = \{A(t); t\geq 0\}$$ and $$B = \{B(t); t\geq 0\}$$ be the forward and backward recurrences time respectively. Find a renewal equation for $$h(t) = P\{A(t) >x, B(t) >y\}.$$
My Take: I am able to figure out that the renewal equation for $A$ is obtained as $$P\{A(t) > x, T_1 \leq t \} = \int_0^t P\{A(t-u)>x\} dF(u)$$ such that the renewal equation is $$h(t)_A = 1 - F(t+x) + F \star h(t).$$
Similarly for $B$ it is obtained as follows,
$$P\{B(t) > x, T_1 \leq t \} = \int_0^t P\{B(t-u)>x\} dF(u)$$ and $$h(t)_B = (t>x) (1-F(t))+ F \star h(t).$$
Thus, I am not sure whether combining $h(t)_A$ and $h(t)_B$ is legit for the joint renewal equation for $A$ and $B$ i.e., $h(t) = P\{A(t) >x, B(t) >y\}.$ I welcome any ideas to this problem. You could provide your solution as well.
 A: Conditioning on the time of the first renewal is the right way to start, but your results don't seem correct. For $s\geqslant 0$ consider
$$
\mathbb P(A(t)>y) = \int_0^\infty \mathbb P(A(t)>y\mid T_1=s)\ \mathsf d F(s).
$$
For $s\in[0,t]$ the integrand above is equal to $\mathbb P(A(t-s)>y)$, for $s\in (t,t+y]$ it is zero, and for $s>t+y$ it is one. Hence
$$
\mathbb P(A(t)>y) = \int_0^t \mathbb P(A(t-s)>y)\ \mathsf d F(s)+\int_{t+y}^\infty \ \mathsf d F(s) = 1-F(t+y)+h_A(y)\star F,$$
where the renewal equation is $h_A(y) = 1-F(t+y)+h_A(y)$.
For the backward recurrence time, note that $\mathbb P(B(t)\geqslant x) = \mathbb P(A(t-x)>x)$ for $x\in[0,t]$, from which it follows that
$$
\mathbb P(B(t)\geqslant x) = 1-F(t+\int_0^{t-x}(1-F(t-s))\ \mathsf d m(s).
$$
Finally, $\mathbb P(B(t)\geqslant x,A(t)>y) = \mathbb P(A_{t-x}>x+y)$. From the above it follows that
$$
\mathbb P(B(t)\geqslant x, A(t)>y)=1-F(t+y)+\int_0^{t-x}(1-(F(t+y-s))\ \mathsf d m(s),\quad 0\leqslant x\leqslant t, y\geqslant 0.
$$
