SOLUTION: Today I asked my Professor in my Logic class if I could do this.
1. ~P
2. ~(PvQ)>~R
3. ~Q ...(conclusion) ~R
4. ~Pv~Q 1,Add
5. ~R 2,1 MP
He flipped
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Proofs
-> SOLUTION: Today I asked my Professor in my Logic class if I could do this.
1. ~P
2. ~(PvQ)>~R
3. ~Q ...(conclusion) ~R
4. ~Pv~Q 1,Add
5. ~R 2,1 MP
He flipped
Log On
Question 1101290: Today I asked my Professor in my Logic class if I could do this.
1. ~P
2. ~(PvQ)>~R
3. ~Q ...(conclusion) ~R
4. ~Pv~Q 1,Add
5. ~R 2,1 MP
He flipped out and asked how in the world did I got this because ~Pv~Q is not the same as ~(PvQ). He very rudely told me I should know this by now. Which I obviously don't and was curious why those are not equivalent and if they ever would be equivalent. Also, what would be the way to solve this proof then? Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! ~P v ~Q translates to "not P or not Q"
so for example, if we have these two statements
P = pirates are real
Q = quiz is tomorrow
then ~P v ~Q would translate to "pirates are not real OR the quiz is not tomorrow"
Now consider the expression ~(P v Q)
That would be "It is not the case that pirates are real or the quiz is tomorrow"
In other words if person A says "pirates are real or the quiz is tomorrow", then person B can counter with the complete opposite and say "pirates are not real AND the quiz is not tomorrow". The "or" flips to "and". We negate everything. Even the connector between the two statements.
This is De Morgan's Law in action. We can use truth tables to see that ~P v ~Q and ~(P v Q) are not equivalent. Compare the columns highlighted in yellow. The truth values don't match up.
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Here is how to solve the proof
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