Here's my shot at Alex Varga's fascinating puzzles, for partially ordered sets (I'm assuming a partial order so that "\$$x AV1: Let's suppose \\(x\leq y\$$. We want to show that \$$(x,x)\leq (y,y)\$$ in the lexicographic order, i.e. \$$x AV2: We wish to find some function \\(r(y,z)\$$ such that \$$x\leq r(y,z)\iff (x,x)\leq (y,z)\$$. Expanding out the latter relation we have "\$$x < y\$$ or (\$$x=y\$$ and \$$x\leq z\$$)". There are two cases: either \$$y\leq z\$$ or \$$y\not\leq z\$$. If \$$y\leq z\$$, then "\$$x < y\$$ or (\$$x=y\$$ and \$$x\leq z\$$)" is equivalent to "\$$x\leq y\$$", so \$$r(y,z) = y\$$. If \$$y\not\leq z\$$, then "\$$x=y\$$ and \$$x\leq z\$$" is false for every \$$x\$$, so "\$$x < y\$$ or (\$$x=y\$$ and \$$x\leq z\$$)" is equivalent to "\$$x \$r(y,z) = \begin{cases}y\text{ if }y\leq z\\\\y'\text{ otherwise.}\end{cases}\$ Otherwise, no such adjoint \\(r\$$ exists.

I imagine AV3 will be similar but I haven't worked it out.