# Analysis of properties of logic operators.

#### Tacticus

Joined Sep 21, 2010
2
The question is attached to the post.

I'm not sure how I should start in this question.

Commutative

xy = yx

Associative

x + y = y + x

so, are they asking me to prove:

(x -> y) = ( y -> x)

and

(x -> y) + (y -> x) = (y -> x) + (x -> y)

This question is really confusing me.

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#### Georacer

Joined Nov 25, 2009
5,182
First of all, get your thoughts right. You have misunderstood the associative property. Look, here for an explanation: http://www.onlinemathlearning.com/number-properties.html (a random find among many).

In order to test the commutative property of the $$\rightarrow$$ operation, put in the first column of the truth table the operand y and x on the second. Apply $$\rightarrow$$ as the truth table shows you and compare the results you got with the third column of the initial table. Are they the same?
For example: 0$$\rightarrow$$1=1. Does 1$$\rightarrow$$0=1? If not, then $$\rightarrow$$ is not commutative.

Same for associative. Take permutations of 3 variables x,y and z, for the operation (x$$\rightarrow$$y)$$\rightarrow$$z. If all the results of this expression are the same with the results of the expression x$$\rightarrow$$(y$$\rightarrow$$z), then the $$\rightarrow$$ operator is associative.