Learning Logic and Truth Tables: My Algebra Journey

[Shackleford]

I'm taking Abstract Algebra right now, and we just briefly covered Logic and Truth Tables. This is my first time in school to learn such things.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110712_195458.jpg

37. I understand.

39. I understand.

41. I don't understand the last column. Is the implication (p ∧ q) ⇒ p true because (p ∧ q) gives you no information on whether p is true? Why?

43. For p ⇒ q, I understand the first two column entries (T,F), is the implication because p being False gives no information on whether p ⇒ q is True? Again, for the last column, is
(p ∧ (p ⇒ q)) True because it being False gives you no new information on q?

If my reasoning is correct, then I can see the following logical consistencies here. Also, I notice for the implications ⇒, there is an additional column asking for its truth value. But, there is no such thing for the iff ⇔ statements. Why?

 

[lanedance]

41 is called vacuously true, see
http://en.wikipedia.org/wiki/Vacuous_truth

basically in a statement
$$P \implies Q$$

if P is false, then the statement is vacuously true, as the rest of the statement no longer needs to be evaluated

consider in the form
if P then Q
when P is false the statement has no more information. In an analogy with programming, there is no elseif or else so nothing else to evaluate making it vacuously true

 

[SammyS]

41. is true because the only way for $A\implies B$ to be false is for A to be true and B to be false.

(p AND q) is always false when p is false.

 

[SammyS]

...

43. For p ⇒ q, I understand the first two column entries (T,F), is the implication because p being False gives no information on whether p ⇒ q is True?

I would word this as: When p is false , then p ⇒ q gives no information as to whether q is true or false.

 

[Mark44]

41. I don't understand the last column. Is the implication (p ∧ q) ⇒ p true because (p ∧ q) gives you no information on whether p is true? Why?

41. is true because the only way for $A\implies B$ to be false is for A to be true and B to be false.

(p AND q) is always false when p is false.

Another way to think of this (#41) is this: If p and q are true, then obviously p is true. There's a related statement -
(p ∧ q) ⇒ q

 

[Shackleford]

Okay. I got it. Thanks for the help, guys.