How does associativity work for this fuzzy norm?

Question

How to prove that $$\min\left(1,\min(1,x+y)+z\right)=\min\left(1,x+\min(1,y+z)\right)$$ where $x,y,z\in[0,1]$?

Answer 1

I don't know what's fuzzy about this.

What would we need for the following statement? $$\min(1, \min(1, x+y) + z) < \min(1, x + \min(1, y+z))\tag{1}$$

Note that $\min(a,b) < c$ is equivalent to $(a < c) \ \text{or}\ (b < c)$, while $a < \min(b,c)$ is equivalent to $(a < b) \ \text{and}\ (a < c)$. Using these a few times, and noting that $1<1$ and $0 < 0$ are false, we find that (1) is equivalent to

$$ ((0 < -z) \; \text{and}\; (-x + z < 0)\; \text{and}\; (1-x < y))\ \text{or}\ ((-x + z < 0) \; \text{and}\; (x + y < 1 - z) \; \text{and} \; (1 - x < y))$$ Both clauses here are incompatible with $z \ge 0$. So we conclude that (1) is impossible if $z \ge 0$.

Since $$\min(1, \min(1, x+y) + z) > \min(1, x + \min(1, y+z))\tag{2}$$ is equivalent to what you get from (1) by interchanging $x$ with $z$, that is incompatible with $x \ge 0$.

We conclude that your equation is always satisfied if $x \ge 0$ and $z \ge 0$.