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

\\((\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