I think we can simplify that slightly @Matthew

We have your (A) and (B), ie \$$x\otimes (x\multimap y) \leq y\$$ and \$$y \otimes (y\multimap z) \leq z\$$.

Therefore \$$x\otimes (x\multimap y) \otimes (y\multimap z) \leq y \otimes (y\multimap z) \leq z\$$.

So by adjunction \$$(x\multimap y) \otimes (y\multimap z) \leq (x\multimap z)\$$.

(EDIT: Oh I see this is the same proof, but you spelled it out in a bit more detail!)