Please help me to Validate the argument by rules of inference
~R
Q => R
P V Q
Then P ^ ~Q

- anonymous

Please help me to Validate the argument by rules of inference
~R
Q => R
P V Q
Then P ^ ~Q

- chestercat

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- anonymous

Rule of inference

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- anonymous

dunno how to put this to words, but with ->, see: http://www.millersville.edu/~bikenaga/math-proof/truth-tables/truth-tables13.png
T F -> F
F F -> T
we have ~R and Q=>R given which means Q is also false
the only way we can have P V Q is if P is true
I'm not quite sure what I'm doing is right

- anonymous

It valid (if solve by truth table) but I try to use by rules of inference, I can't solve
i wanna know about solution to solve it

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## More answers

- anonymous

~R
Q => R
is modus tollens

- anonymous

so now we have
modus tollens
~R
Q => R
.:.~Q
-----------------
P V Q
Then P ^ ~Q
---------------------------
with elimination, we have
~Q
P V Q
.:. P

- anonymous

sorry that's so disorganized

- anonymous

and what we will do with P ^ ~Q, how I know it is true

- anonymous

I don't think there's an argument for it, it's just what we found
we got P from elimination and ~Q from modus tolens

- anonymous

Then we can use P and ~Q with conjuction ?

- anonymous

oh, wow I'm blind
yeah

- anonymous

I feel confuse about definition of Propositional Logic that can use ~Q again

- anonymous

do you understand why modus tollens is true logically?

- anonymous

I dunno I use follow the rule

- anonymous

no point in following rules blindly. this is the truth table we're working with\(
\begin{array}{l|c|r}
\text{P} & \text{Q} & \text{P}\implies \text{Q}\\
\hline
0 & 0 & 1 \\
0 & 1 & 1\\
1 & 0 & 0 \\
1 & 1 & 1
\end{array}
\)

- anonymous

modus tollens states that \(
p\implies q\\ \text{~}q\\\text{.^.~p}
\)

- anonymous

I think I have more understand it
same modus tollens that you said from p => q is T
and ~q is T then q = F
then if p => q will T and p = F then ~p = T

- anonymous

when we're given a statement, we assume it's true
the ~ is a negation
so p=>q means we only look at the cases where p=>q evaluates to true

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- anonymous

then we look at the next statement ~q
we need to find a place where q is 0

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- anonymous

the conclusion from modus tollens says we have ~p or p=false

- anonymous

and we can see it's true in the truth table

- anonymous

Thanks a lot!
I quite understand it

- anonymous

no problem, what class is this for btw

- anonymous

Discrete Math

- anonymous

ah, same. I assume you're a fellow CS major

- anonymous

out of curiosity, how much calc did you have to take

- anonymous

I have grade B calc since freshman until second year

- anonymous

I'm sorry if i slow to reply
I'm not strong in English language, I try to use it more

- anonymous

that's the best way to learn a language :)
anyhow good luck with your class

- anonymous

You too, Good luck
and Thanks for everything that you gave me today.

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