The smallest $T$-norm

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 .

The drastic product ($T(x,y) = \begin{cases} \min(x,y), & \text{if $\max(x,y)=1$} \\ 0, & \text{otherwise} \end{cases}$) clearly is a T-norm

Now show that the drastic product is the least T-norm .

Answer 1

Consider $T'$ satisfying $T_1-T_4.$ Then, if $\max\{x,y\}<1$ it is clear that $$ T(x,y)=0\le T'(x,y).$$ And if $\max\{x,y\}=1$ (assume $x<y=1$ without lost of generality, since $T(x,1)=T(1,x)$) we have that, because of $T_1,$

$$T(x,1)=x=T'(x,1).$$

Thus, we have shown that the drastic product is the smallest $T$-norm.