Analysis of properties of logic operators.

Thread Starter

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.

Thanks for help in advance.
 

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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.
 
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