For two random variables $X,Y$ and two constants $a>b$ I wonder whether the following set of equations holds. $$ P(X<a-Y|b<X) = P(b<X<a-Y) = P(b<a-Y)$$
In the first equation I wonder whether I may change the conditioning into an ordered probability. In the second question I wonder whether I may evaluate only the outer ordering to arrive at the same probability; the latter I found in a proof and doubt it.
Neither of your equations holds in general. Since $$ P(X<a-Y\mid b<X) = \frac{P(b<X<a-Y)}{P(b<X)}\ , $$ the first equation will hold if and only if either $\ P(b<X)=1\ $ or $\ P(b<X<a-Y)=0\ $, and it is easy to find examples where neither of those conditions is satisfied.
For an example in which none of the three probabilities is equal to either of the other two, take $\ X\ $ and $\ Y\ $ independent and both uniformly distributed over the interval $\ [0,1]\ $, and $\ b=\frac{1}{2}, a=1\ $. Then \begin{align} P(X<a-Y\mid b<X)&=\frac{1}{4}\ ,\\ P(b<X<a-Y)&=\frac{1}{8}\ ,\ \text{and}\\ P(b<a-Y)&=\frac{1}{2}\ . \end{align}