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- genericname
- 2012-10-17 06:31:21
Thanks, Bob. 
- bob bundy
- 2012-10-17 05:41:12
hi genericname
The truth table below shows what I did. I'm not saying it is correct; to tell the truth that comma has me puzzled too. I'd prefer all compound statements to be made using only propostional calculus notation.
Have a look back at your notes / text book. What is a comma supposed to mean?
If you take a look at that link I gave you, you'll find the author uses two different notations:
notation 1
statement 1
statement 2 underneath
notation 2
statement 1 & statement 2
I assumed that the comma was the equivalent of notation 1.
Bob
- genericname
- 2012-10-17 05:27:11
bob bundy wrote:
hi genericname
The notation is a little unclear but compare with
http://www.cs.odu.edu/~toida/nerzic/lev … L/E2L.html
I think
p, p<=>q means p and p<=>q
The truth tables agree with p^q and you can only test the equivalence if you can also consider the cases where p is false.
Hope that's right.
Bob
So it saying to compare the values of p^q with p<=>q?
p | q | p<=>q | p^q |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 |
- bob bundy
- 2012-10-17 05:17:36
Do you know what a truth table is?
Bob
- anonimnystefy
- 2012-10-17 05:01:42
bob bundy wrote:
The truth tables agree with p^q and you can only test the equivalence if you can also consider the cases where p is false.
What do you mean by this?
- bob bundy
- 2012-10-17 04:30:21
hi genericname
The notation is a little unclear but compare with
http://www.cs.odu.edu/~toida/nerzic/lev … L/E2L.html
I think
p, p<=>q means p and p<=>q
The truth tables agree with p^q and you can only test the equivalence if you can also consider the cases where p is false.
Hope that's right.
Bob
- genericname
- 2012-10-17 04:10:00
http://i.imgur.com/qGLc1.png
Hi,
Does it mean that all the P's for that problem are true? The problem is telling me to use a truth table to tell whether or not the statement is true and the 'p' in front put me off.