Logic: Prove validity of Modus Ponens argument

syno

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Hi. I am trying to prove the validity of this argument:

P ⇒ ((∼Q) ∧ R)
Q ⇒ (P ∨ R)
therefore P ⇒ R

I have constructed truth tables for each statement, however I have not need Modus Ponens beyond it's most basic form before and I'm really confused on how to proceed to determine the answer.
 
I am perplexed. Your first line alone is sufficient to deduce your third line.

[MATH]P \implies (\neg Q) \land R) \\ P \implies R.[/MATH]
If X is a dog, then X is not a snake and X has legs means that if X is a dog, it has legs.

And what does any of this have to do with modus ponens, which goes

[MATH]J \implies K \\ J \\ \therefore K.[/MATH]
Perhaps you are trying to go

[MATH]P \implies \neg Q \land R \\ and \ Q \implies P \lor R \\ \therefore Q \implies R.[/MATH]
I do not have a name for that argument, but it looks like it uses modus tollens

[MATH]\{P \implies \neg Q \land R\} \land \{Q \implies P \lor R\}.\\ \\ \{P \implies \neg Q \land R\} \implies \\ \{\neg \{\neg Q \land R\} \implies \neg P\} \implies \\ \{Q \lor \neg R \implies \neg P\} \implies \\ \{Q \implies \neg P\} \\ But \ \{Q \implies P \lor R\} \\ \therefore Q \implies R.[/MATH]
If X is a dog, then X is not a snake and X has legs
If X is a snake or X has no legs (whether because X is a snake or because X is an oyster), then X is not a dog.

If X is a snake, then X is a dog or X has no legs. True, but we can simplify to if X is a snake, then X has no legs.

That argument seems to be a very complicated way of saying

If A entails B or C
and
If A entails not B,
then A entails C.
 
Last edited:
Hi. I am trying to prove the validity of this argument:
P ⇒ ((∼Q) ∧ R)
Q ⇒ (P ∨ R)
therefore P ⇒ R
Here is what I would do.
\(p\Rightarrow(\neg Q\wedge R)\)
\(\neg P \vee(\neg Q\wedge R)\)
\((\neg P \vee\neg Q)\wedge(\neg P \vee R)\)
\((\neg P \vee R)\)
\(P\Rightarrow R\)
 
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