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I'm looking at proving that the Łukasiewicz T-norm operator, or bounded product T-norm, is a T-norm, but I'm stuck on associativity.

The operator is defined as:

$xTy = max[0,x+y-1]$

Trying to show associativity:

$(aTb)Tc =$

$max[0,max[0,a+b-1]+c-1] =$

$max[0,max[c-1, a+b+c-2]] =$

$max[0,max[c-a,b+c-1]+a-1] =$

$max[0,a+max[c-a,b+c-1]-1] =$

At this point, I need to show that this is equivalent to...

$=max[0,a+max[0,b+c-1]-1]=aT(bTc)/$

But I can't figure out how to resolve $c-a$ to zero.

Hopefully someone can help!

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    Hint: if either $aTb$ or $bTc$ are 0, you can show both sides of your equation equal 0. If both $aTb$ and $bTc$ are positive, then there is one algebraic step to go from what you have to what you want. Hope this helps. If you are still stuck I will put more details in an answer. – 2017-02-05

1 Answers 1

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Picking up from where you have $max[0,max[0,a+b−1]+c−1]$, there are two possibilities for the inner $max$. It either evaluates to $a+b−1$ or $0$.

In the first case we will have:

$$ max[0,max[0,a+b−1]+c−1] = max[0, a+b+c-2] $$

And in the second case we have: $$ max[0,max[0,a+b−1]+c−1] = max[0, c-1] $$

But $c-1 \in [-1, 0]$ since $c \in [0, 1]$. So we have: $$ max[0, c-1] = 0 $$

If you do the same for your other equation:

$$ aT(bTc) = \begin{cases} max[0, a+b+c-2], \text{ when }(bTc) \neq 0 \\ max[0, a-1] = 0, \text{ otherwise} \end{cases} $$