anonymous
  • anonymous
Please help me to Validate the argument by rules of inference ~R Q => R P V Q Then P ^ ~Q
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Rule of inference
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anonymous
  • 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
  • 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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anonymous
  • anonymous
~R Q => R is modus tollens
anonymous
  • 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
  • anonymous
sorry that's so disorganized
anonymous
  • anonymous
and what we will do with P ^ ~Q, how I know it is true
anonymous
  • 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
  • anonymous
Then we can use P and ~Q with conjuction ?
anonymous
  • anonymous
oh, wow I'm blind yeah
anonymous
  • anonymous
I feel confuse about definition of Propositional Logic that can use ~Q again
anonymous
  • anonymous
do you understand why modus tollens is true logically?
anonymous
  • anonymous
I dunno I use follow the rule
anonymous
  • 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
  • anonymous
modus tollens states that \( p\implies q\\ \text{~}q\\\text{.^.~p} \)
anonymous
  • 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
  • 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
  • anonymous
then we look at the next statement ~q we need to find a place where q is 0
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anonymous
  • anonymous
the conclusion from modus tollens says we have ~p or p=false
anonymous
  • anonymous
and we can see it's true in the truth table
anonymous
  • anonymous
Thanks a lot! I quite understand it
anonymous
  • anonymous
no problem, what class is this for btw
anonymous
  • anonymous
Discrete Math
anonymous
  • anonymous
ah, same. I assume you're a fellow CS major
anonymous
  • anonymous
out of curiosity, how much calc did you have to take
anonymous
  • anonymous
I have grade B calc since freshman until second year
anonymous
  • anonymous
I'm sorry if i slow to reply I'm not strong in English language, I try to use it more
anonymous
  • anonymous
that's the best way to learn a language :) anyhow good luck with your class
anonymous
  • anonymous
You too, Good luck and Thanks for everything that you gave me today.

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