Can't Prove $T(x,0)=0$.

Question

Let $T : [0, 1]\times [0, 1] \rightarrow [0, 1]$. A $t$-norm is a function $T$ with properties:

$1. T (x, 1) = x$

$2.$ If $y\leq z$ then $T(x,y)\leq T(x,z)$

$3. T (x, y) = T (y, x) $

$4. T (x, T (y, z)) = T (T (x, y), z) $

Prove $T(x,0)=0$.

I'm trying to prove $T(x,0)\geq 0$ and $T(x,0)\leq 0$.

Obviously, $T(x,0)\geq 0$ because the definition of $T$.

But I can't prove $T(x,0)\leq 0$.

Any hint to prove it?

Answer 1

We have $T(0,1)=0$. Hence $0 \leq T(x,0)=T(0,x) \leq T(0,1)=0$.

I hope you can see what properties I have used in each step of the argument.