calculate probabilty, Uniform distribution

Question

this is my first question so excuse my unknowing and mistakes: I was reading a book and just faced this thing:

(1.4) $=P(X\gt Z/2)(Y-X)$

(1.5) $=P(2X\gt Z)(Y-X)$

(1.6) $=\min\{{2X,1\}}(Y-X)$

I'm facing difficulty in understanding the transition from equation (1.5) to (1.6). what is given is that $X$ and $Z$ are uniformly distributed between $[0,1]$.

Answer 1

If $2X>1$ then $P(2X>Z)=1$. If $2X<1$ then $P(2X>Z)=2X$, since $Z$ is uniform on $[0,1]$.

Answer 2

The proof asserts: $P(2X\gt Z) = \min\{{2X,1\}}$

and the OP asks why this is so. Neither the proof nor the answer above make any sense to me.

If X ~ Uniform(0,1) and Z ~ Uniform(0,1) are independent, then $P(2X\gt Z)$ = $\frac{3}{4}$.
By contrast, $\min\{{2X,1\}}$ = 2X if x < $\frac{1}{2}$, and 1 otherwise. It is not a constant, nor a probability: it is a random variable.