Looks like the third one is true as well

\$$(\hat{F} \circ \hat{G})(x, z)\$$

\$$= \bigvee\big(\hat{G}(x, y) \otimes \hat{F}(y, z)\big)\$$

\$$= \bigvee\big(Y(Gx, y) \otimes Z(Fy, z)\big)\$$

And it turns out this join \$$= Z(FGx, z) = \widehat{FG}(x, z)\$$

To prove this, first set \$$y = Gx\$$ to see that our join

\$$\geq Y(Gx, Gx) \otimes Z(FGx, z) \geq I \otimes Z(FGx, z) = Z(FGx, z)\$$

But given any \$$y\$$ we have

\$$Z(FGx, z) \geq Z(FGx, Fy) \otimes Z(Fy, z) \geq Y(Gx, y) \otimes Z(Fy, z)\$$

So we also have \$$Z(FGx, z) \geq\$$ our join. QED