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.

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

Otherwise, no such adjoint \\(r\\) exists.

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