Question

Truth Table P ---> Q clarification.

We all know that the truth value of P is T T F F and the truth value of Q is T F T F, but what I don't understand is why the final result is true if P is false and Q is true.

Answer 1

Suppose I said "if it rains, it is cloudy"

Suppose it doesn't rain, but it is cloudy.
My statement is still true in this case.

Answer 2

what about this sentence for an example...
If I get an A, then I'll get a Pizza.
P:I get an A
Q: I'll get a pizza

So If P is false which means I don't get an A but Q is true which is I'll get a pizza, why on earth is that still true for P ---> Q?

Answer 3

Well, because maybe you saved Papa John, and he gave you a pizza for that reason.

Answer 4

XD oh by the way, I found extra problems from previous sections which was identical to the p diamond q problem. da...mn if only that was assigned earlier, I would've known instead of going nuts for three hours. oy xx,,,xx

Answer 5

lol they did that using truth tables is it ?

Answer 6

geneshie, where you get that picture?

Answer 7

for ur original question,

try to *compare* it with biconditional, and see why p-->q and p<-->q are different :
I'll get a Pizza 'if and only if' I get an A

Answer 8

my best friend sketched it lol xD

Answer 9

@ganeshie8 yeah only it was a P CIRCLE PLUS Q

Answer 10

plus in the circle XD

Answer 11

+ in the circle is XOR

'either this or that', but not both.

Answer 12

the p <--> Q makes a lot of sense... since the original condition was If I get an A, then I'll get a pizza.
P is false Q is True P<---> Q
If I don't get an A, then I'll get a pizza breaks the deal. so F!

Answer 13

xor is exclusive... had its own truth table.

Answer 14

but it wasn't assigned... the p diamond q was...maybe that's why I went coo coo. the equilvalent problems were easy probably they were assigned before that's how i got the first 2 fast

Answer 15

If I get an A, then I'll get a Pizza.

see if below looks plausible.
we dont know if there are other ways to get a Pizza; so premise is not fully known here. so when P is false, we *simply* take the outcome as true.

Answer 16

then another problem in the book was prove that <-> is commutative and associative... wouldn't that be P <---> Q q <---> P for commutative
and P <-->( Q <---> R) (P <---->Q) <----> R

Answer 17

yes ! looks you're getting good at these lol

Answer 18

yeah I've reread the previous sections and did the problems that weren't assigned as practice. Too bad there's no answer sheet in the back of the book. pure proofs = NO ANSWER 4 U D://// but how will I know if I'm right D:

Answer 19

the discrete book I downloaded with the manual has plenty of problems and goes into detail

Answer 20

Easiest way to prove is with truth tables. If that is not available, or there are too many variables, then you use algebra.

Answer 21

check it wid wolfram, if you wry about making mistakes in truth tables / algebra

Answer 22

ok but what if there is no example for the algebra version?

Answer 23

that's how I got easily lost when I tried to do 1.4.15

Answer 24

wolfram is very good in logic, it gives u nice truth tables also

Answer 25

can wolfram do the algebra version .. like if ~(~P) <-> P
P <-> P because of double negation law... can wolfram understand that or no?

Answer 26

http://www.wolframalpha.com/input/?i=~%28~P%29+%3C-%3E+P

Answer 27

nothing much to do for wolfram... its trivial

Answer 28

http://www.wolframalpha.com/input/?i=%28~%28~P%29%29+%3C-%3E+P

Answer 29

looks negation has precedence over <->, so careful wid these... always add parenthesis..

Answer 30

i like this site for generating truth tables for any given function :-
http://turner.faculty.swau.edu/mathematics/materialslibrary/truth/