logicians, check this out
 

Question 13
Given that p is true, q is false, and r is true, complete the one line truth table for:


p q r (p → ~p) V r


p q r (p → ~p) V r

T F T F T F T T


p q r (p → ~p) V r

T F T F F F T T


p q r (p → ~p) V r

T F T F T F F T


p q r (p → ~p) V r

T F T F T F F T
Question 14

Given that p is true, q is false, and r is true, complete the one line truth table for:

~r ↔ (q V p)


p q r ~ r ↔ (q V p)

T F T F T T T F


p q r ~ r ↔ (q V p)

T F T F F T F F


p q r ~ r ↔ (q V p)

T F T F T T F F


p q r ~ r ↔ (q V p)

T F T F F T T F



if you know your logic you would see an issue here... I think, I am most certain these questions are wrong since if you see the cases do not stay the same even though they are established prior to the problem.

the answers are below the problems

p q r (p → ~p) V r

That's just silly. p = true always means ~p = false, so that equation is false. Since the next operator is a logical disjunction, and r is true, the result of the entire equation is true.

~r ↔ (q V p)

since r is true, ~r is false.
since p is true, q V p is true
false ↔ true = true
so the result is true.

The rest is just semantics.

Dang, its been a long time since I studied logic.

I think I have forgotten most of what I learned then.

(a -> b) is equivalent to (~a V b). So (p -> ~p) is equivalent to (~p V ~p) which is simply ~p. Is that correct?

The truth values are incorrect.. its just a one level truth table with the truth values given... where the p is listed as false in the first problem its defined as True!!!! regardless.. its a simple problem that should not be over complicated by any outside forces.. Am I correct?

:shocked2:

Here is how you do it. First you have these rules.

P -> Q is false only once, when P is True and Q is false.

P & Q is true only once, when both P and Q are true.

P v Q is false only once, when both P and Q are false, or, only one variable needs to be true in order for the statement to be true.

P <-> Q is true only when both P, Q, have the same values. ie., when P and Q are True then the <-> is true.

So first, fill in underneath your P, Q, and R, your given values. So

Then you fill in the values for the operators -> and v.

So

p -> ~p is false, so fill in F under the conditional arrow.

r is true, so as per the rule, that a disjunction only needs one variable true to be true, fill in T underneath the v sign.

So the statement is True.