Question

see attachment for questions. I know the first two, but for the last two I have to prove that my assumptions are correct...one was obvious and the other was not.

Answer 1

@LastDayWork @ganeshie8 @phi

Answer 2

hmmm I wonder if I should create a truth table for part c and compare it to the p diamond q . I know the associative law isn't going to work because we are given two statements of p and q, not p q and r

Answer 3

??

Answer 4

like should the truth table be allowed the first part of the communtative law will have T F F F
and the second will be T T T F
taking the AND into effect and it will be T F F F which is no where near p diamond q...but there's got to be something else besides that

Answer 5

Consider <diamond> as any other binary operator (like + or *)

So, do you still think "...I know the associative law isn't going to work because we are given two statements of p and q, not p q and r..." ??

Answer 6

<> is commutative iff P <> Q = Q <> P

Answer 7

So the most easiest approach would be to make a truth table -
P Q P <> Q Q <> P

Don't use part (a) or (b) to solve (c)

Answer 8

I think part a and b are separate... those were easy to do... I did have a truth table for c ...

Answer 9

P is T T F F
Q is T F T F
??? how can P and Q be equivalent in this?

Answer 10

What do you mean by "...how can P and Q be equivalent in this?..." ??

Answer 11

ok wait I've stated the law in the attachment
P and Q has the truth table of T F F F
Q and P has the truth table of T F F F ok same truth table value equilvalent

P or Q T T T F
Q or P T T T F

Answer 12

the question is asking if P DIAMOND Q is commutative. I've stated the law in the attachment
P DIAMOND Q has the truth table of T T F T

Answer 13

so the question is is the diamond commutative ? not sure how to put this in words
is the diamond associative? doubt it. we need three statements and the diamond only has p and q

Answer 14

Consider the following truth table