I got the following proof wrong in class and I could use some help solving it:


-(-P v -Q) therefore (P & Q)

(Moderator edit: post moved to more appropriate forum.jgw)

Kennethamy is right in that it is a replacement rule for DeMorgans. But there is a little problem with it.

Your translation of DeMorans looks like this;
-(-P v -Q) therefore (P & Q)

The ~(~P v ~Q) is correct, this is an acceptable starting point for DeMorgans. However the issue lies in the replacement formula that follows.

DeMogans goes as follows.

~(P v Q) may replace or be replaced by ~P & ~Q
~(P & Q) may replace or be replaced by ~P v ~Q

So whenever you want to use DeMorgans law, you have to exchange the ampersand with the wedge or the wedge with the ampersand, negate each side of of the ampersand or wedge, and negate the formula as a whole.

So the correct replacement should have been:
-(-P v -Q, )therefore -P & -Q

I got the following proof wrong in class and I could use some help solving it:


-(-P v -Q) therefore (P & Q)

(Moderator edit: post moved to more appropriate forum.jgw)

DeMorgans allows us to derive something bi-equivalent with what has been replaced, hence it is a replacement rule. DeMorgans is basically the same thing as saying something which is the same topic, but in a different way.

So take what I had said was DeMorgans rule, namely ~(P&Q) is equivalent to ~Pv~Q. Translated into syntactics, "Alan and Bill are both not at home" [i.e. ~(P&Q)] is the same as saying "Either Alan is not home or Bill is not at home" [i.e. ~P v ~Q]

If we do change it the way you suggest, namely change the sign and remove all the negations, we will end up with a non-equivalent replacement. When translated, we come up with P v Q, the same as saying "Either Alan or Bill are not at home."

Is "Alan and Bill are both not at home" the same as saying "Either Alan or Bill are not at home?"

Is "Alan and Bill are both not at home" the same as saying "Either Alan or Bill are not at home?"

Isn't that saying two different things? In the first sentence both Alan and Bill are not home, but in the second it suggests that either one of the two are not home, leaving open the option that Alan or Bill could be potentially at home.

DeMogans goes as follows.

~(P v Q) may replace or be replaced by ~P & ~Q
~(P & Q) may replace or be replaced by ~P v ~Q

So whenever you want to use DeMorgans law, you have to exchange the ampersand with the wedge or the wedge with the ampersand, negate each side of of the ampersand or wedge, and negate the formula as a whole

So the correct replacement should have been:
-(-P v -Q, )therefore -P & -Q

-(-P v -Q, )therefore -P & -Q

as you wrote in your post for the correct replacement:


let's just test this using a truth table:

~P ~Q (~P v ~Q) ~(~P v ~Q) ~P&~Q P Q P&Q
T T T F T F F F
T F T F F F T F
F T T F F T F F
F F F T F T T T

how can we replace ~(~P v ~Q) with ~P&~Q if they have different truth values?

Kata,

The truth table that you are formulating is not correct. Individual truth tables have to be developed in order find the truth value of each permutation of DeMorgans. When they are done correctly, they look something like this.

In the ~(PvQ) truth table, all of the formal truth value translations have been done. The negation is the primary symbol to base our general truth value for the table. The rule for a disjunction states that a "disjunction is false when both disjuncts are false." With this done and then negated, the general truth value is F,F,F,T. In the ~P&~Q, we extrapolate the truth values and apply the law "a conjunction is true only if both conjuncts are true" and come up with exactly the same truth values as the bi-equivalent translation ~(PvQ).

So in response to your question... we can replace them becuase they have the same truth value.


Kennethamy,

At what point are you using P&Q as a replacement?