Propositional logic - Tilted Forum Project Discussion Community

Ace_O_Spades · 01-22-2004, 06:34 PM

ok i know there's a lot of different notations for logic, but here's what im gonna use for this question

<=> logically equivalent to
v or
^ and

prove:

p <=> p^(p v q)

this question, while pegged as being easy, has really got my goat right now

KnifeMissile · 01-22-2004, 07:45 PM

First, we can tell by inspection that (p V q) is equivalent to p for all values of p and q except when p is false and q is true. So, because (p ^ p) is equivalent to p, we can immediately eliminate all but one combination and that's when p is false and q is true. So, we can simply inspect that one case and see if they are logically equivalent. Of course, this problem is so easy that in the time it took me to explain this logic in English, I could have shown you all four states in a truth table. So, why don't I do that anyways...

Code:

true  <=> true  ^ (true  V true ) <=> true
true  <=> true  ^ (true  V false) <=> true
false <=> false ^ (false V true ) <=> false
false <=> false ^ (false V false) <=> false

As you can see, the two statements are logically equivalent.
QED.

Ace_O_Spades · 01-22-2004, 07:47 PM

AAAHHHHH i hate it when proofs just jump out at you like that, thank you very very much

it makes so much sense now... maybe its time to take a break from propositional logic for the night

Ace_O_Spades · 01-23-2004, 01:16 PM

found a solution using boolean algebra, if anyone is interested... thought i'd share

Using Boolean algebra.

p = p*(p+q)
= p + pq
= p*1 + pq
= p(1+q)
= p(1)
= p

QED.

KnifeMissile · 01-23-2004, 01:58 PM

Yeah, I wasn't sure if you would accept boolean algebra without proof and, well, how would you prove those? Using truth tables, of course!

Ace_O_Spades · 01-23-2004, 02:03 PM

haha truth tables solve everything

too bad my prof doesnt accept truth tables

its all about Laws of Logic and boolean algebra

01-22-2004, 06:34 PM	#1 (permalink)
Ace_O_Spades The Death Card Location: EH!?!?	Propositional logic ok i know there's a lot of different notations for logic, but here's what im gonna use for this question <=> logically equivalent to v or ^ and prove: p <=> p^(p v q) this question, while pegged as being easy, has really got my goat right now

01-22-2004, 07:45 PM	#2 (permalink)
KnifeMissile Location: Waterloo, Ontario	First, we can tell by inspection that (p V q) is equivalent to p for all values of p and q except when p is false and q is true. So, because (p ^ p) is equivalent to p, we can immediately eliminate all but one combination and that's when p is false and q is true. So, we can simply inspect that one case and see if they are logically equivalent. Of course, this problem is so easy that in the time it took me to explain this logic in English, I could have shown you all four states in a truth table. So, why don't I do that anyways... Code: true <=> true ^ (true V true ) <=> true true <=> true ^ (true V false) <=> true false <=> false ^ (false V true ) <=> false false <=> false ^ (false V false) <=> false As you can see, the two statements are logically equivalent. QED.

01-22-2004, 07:47 PM	#3 (permalink)
Ace_O_Spades The Death Card Location: EH!?!?	AAAHHHHH i hate it when proofs just jump out at you like that, thank you very very much it makes so much sense now... maybe its time to take a break from propositional logic for the night __________________ Feh.

01-23-2004, 01:16 PM	#4 (permalink)
Ace_O_Spades The Death Card Location: EH!?!?	found a solution using boolean algebra, if anyone is interested... thought i'd share Using Boolean algebra. p = p(p+q) = p + pq = p1 + pq = p(1+q) = p(1) = p QED. __________________ Feh.

01-23-2004, 02:03 PM	#6 (permalink)
Ace_O_Spades The Death Card Location: EH!?!?	haha truth tables solve everything too bad my prof doesnt accept truth tables its all about Laws of Logic and boolean algebra __________________ Feh.

01-23-2004, 01:58 PM	#5 (permalink)
KnifeMissile Location: Waterloo, Ontario	Yeah, I wasn't sure if you would accept boolean algebra without proof and, well, how would you prove those? Using truth tables, of course!