I apologise for the bad formatting.
Motivation:
In my research, I have developed some ideas to do with semilattices. I regard them each as a set $L$ under an associative, commutative, idempotent binary operation; the distinction between meet and join semilattices is not yet relevant - they are isomorphic in the theory.
The diagram $(1)$ below was found in a picture somewhere in some search of semilattices, and trying to determine whether it is the Hasse diagram of a semilattice has given me a headache. I will attempt to explain why.
The Question:
Consider the directed graph $\require{AMScd}$ $$\begin{CD} c & & d\\ @A t AA \rlap{{_{\rlap{x}}\style{display: inline-block; transform: rotate(30deg)}{{\xleftarrow[\rule{2em}{0em}]{}}}}}{\style{display: inline-block; transform: rotate(150deg)}{{\xleftarrow[\rule{2em}{0em}]{}}}}{_{\llap{y}}} @AA z A\\ a & & b \end{CD}\tag{1}$$
Does $(1)$ define a semilattice (in the sense of a Hasse diagram?
Thoughts:
Is the join $ab$ of $a$ and $b$ defined? Because both $c$ and $d$ are each greater than $a$ and $b$ (via $\{t,y\}$ for $c$ and $\{ x,z\}$ for $d$) and have no intermediaries (so to speak), yet they are distinct. If $\{ a,b,c,d\}$ were a semilattice, then, my confusion has it, $c=ab=d$ while $c\neq d$, a contradiction to the fact that concatenation here is a binary operation!
Yet the source (I lost) has it that it does define a semilattice.
Perhaps I should trust random documents I can't find again less than my poor intuition in the area. But maybe my proof is wrong . . .
Where do I think it's wrong?
My guess is the step $c=ab=d$. Does it make sense to speak of the join? This feels like a restatement of the contradiction of the proof though . . .