A join-semilattice with greatest element is an algebra $(S,\vee, 1)$ of type $(2,0)$ such that $\vee$ is idempotent, commutative, and associative, and $a\vee 1=1$ for all $a\in A$.
Now, let $(A,m,1)$ be an algebra of type $(3,0)$ such that $m$ satisfies the following three identities:
(1) $m(x,y,x)=x$
(2) $m(x,x,y)=m(y,y,x)$
(3) $m[\, m(x,x,y)\,,\, m(x,x,y)\,,\, z\,]=m[\,x\,,\,x\,,\,m(y,y,z)\,]$
for all $x,y,z\in A$.
Define $x\vee y:= m(x,x,y)$ for $x,y\in A$. The claim is that $(A,\vee, 1)$ is a join-semilattice with greatest element.
Certainly $(A, \vee, 1)$ is of type $(2,0)$. Idempotence, commutativity, and associativity of $\vee$ follows from (1), (2), and (3), respectively.
However, I can't figure out why $x\vee 1=1$. Obviously I just need the right combination of (1), (2), and (3), but I'm not seeing it.
The best I can do right now is: \begin{align*}x\vee 1 = m(x,x,1) &= m\bigl[\,m(x,x,x)\,,\, m(x,x,x)\,,\, 1\,\bigr]\tag{1}\\ &=m\bigl[\,x\,,\,x\,,\,m(x,x,1)\,\bigr]\tag{3}\\ &=x\vee m(x,x,1)\,. \end{align*}
For reference, the claim is from this paper: last paragraph on p.129.
Thank you!
As it was indicated in the comments, I'll post an answer to this question.
The claim is false. This will not hold unless you have some equation relating $1$ to $m$, otherwise there is no reason that this constant will be interpreted as the top element in the semilattice. More precisely, if $(A,\vee)$ (with $\vee$ defined as above) has at least two elements, you may interpret $1$ as any element different from the maximum (if it exists).
For an example, take $A:=\mathbb{Z}$ and $m(x,y,z):=\max\{\min\{x,y\},z\}$. It can be easily verified that this choice satisfies the three identities, and certainly $x\vee y= \max\{x,y\}$ is a semilattice operation. But $A$ is not bounded so actually there is no way of interpreting $1$ as the top (or, by the same token, bottom) element of the semilattice.
Finally, notice that in this paper, there is one more identity $m(x,x,1)=1 $ (which is exactly what you need). It seems fairly probable that they meant to include it in the list and simply forgot.