Can you help to prove this property?

Question

Let $T : [0, 1]^2 → [0, 1]$. Consider the following properties:

$T_1 : T(x, 1) = x$

$T_2 : T(x, y) = T (y, x) $

$T_3 : T(x, T (y, z)) = T(T (x, y), z) $

$T_4 $: If $x ≤ u , y ≤ v \Rightarrow T(x, y) ≤ T(u, v) $

Function $T: [0, 1]^2 → [0, 1]$ that satisfies $T_1 −T_4$ is T-norm.

consider $$T_L(x,y) = max(x+y-1,0) $$ we can show that $T_L(x,y)$ is a T norm.

Conditions 1, 2 and 4 are obviously. Can you help me to prove condition 3?

Answer 1

We have that

\begin{align} T_L(x,T_L(y,z))=\max\{x+T_L(y,z)-1,0\}&=\max\big\{x+\max\{y+z-1,0\}-1,0\big\}\\ &=\max\{x+y+z-2,x-1,0\} \end{align}

Now, notice that $x-1\leq 0$, because $x\in[0,1]$. It follows that

$$T_L(x,T_L(y,z))=\max\{x+y+z-2,0\}$$

and similarly for $T_L(T_L(x,y),z)$.