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Mathematics 22 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

Rule of inference

OpenStudy (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

OpenStudy (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

OpenStudy (anonymous):

~R Q => R is modus tollens

OpenStudy (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

OpenStudy (anonymous):

sorry that's so disorganized

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

Then we can use P and ~Q with conjuction ?

OpenStudy (anonymous):

oh, wow I'm blind yeah

OpenStudy (anonymous):

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

OpenStudy (anonymous):

do you understand why modus tollens is true logically?

OpenStudy (anonymous):

I dunno I use follow the rule

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Thanks a lot! I quite understand it

OpenStudy (anonymous):

no problem, what class is this for btw

OpenStudy (anonymous):

Discrete Math

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I have grade B calc since freshman until second year

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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