# Analysis of properties of logic operators.

Discussion in 'Homework Help' started by Tacticus, Oct 9, 2010.

1. ### Tacticus Thread Starter New Member

Sep 21, 2010
2
0
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.

File size:
44.7 KB
Views:
27
2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
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.

3. ### Tacticus Thread Starter New Member

Sep 21, 2010
2
0
Thanks for the help