Is This Logical Argument Valid?

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SUMMARY

The logical argument presented in the discussion involves determining the validity of the statement using Modus Ponens, disjunctive syllogism, and double negation. The conclusion reached is that the argument is valid by transforming the implication into its equivalent disjunction form, specifically using the replacement rule where p → ¬q is equivalent to ¬p ∨ ¬q. The application of double negation (¬¬q ≡ q) and disjunctive syllogism (DS) clarifies the path to the conclusion ¬p.

PREREQUISITES
  • Understanding of logical implications and equivalences, specifically p → ¬q and its transformation.
  • Familiarity with rules of inference such as Modus Ponens and disjunctive syllogism.
  • Knowledge of double negation elimination (DNE) in propositional logic.
  • Basic proficiency in symbolic logic notation and operations.
NEXT STEPS
  • Study the application of Modus Ponens in various logical arguments.
  • Research disjunctive syllogism and its role in logical deductions.
  • Explore double negation elimination (DNE) and its implications in propositional logic.
  • Practice transforming logical implications into their equivalent disjunctions.
USEFUL FOR

Students of logic, educators teaching propositional logic, and anyone interested in enhancing their understanding of logical argument validity and inference rules.

Euler2718
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Homework Statement



Determine whether the following is valid:

p \rightarrow \neg q , r \rightarrow q , r, \vdash \neg p

Homework Equations



Modus Ponens, disjunctive syllogism, double negation.

The Attempt at a Solution



I've boiled it down to

p \rightarrow \neg q , q, \vdash \neg p

However I do not understand how the book says to use disjunctive syllogism and double negation here. I've expanded the implication into its fundamental forum but I still can't see it.
 
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These problems always depend on what rules of inference, axioms and replacement rules you've been given.
##p\to \neg q## is equivalent to ##\neg p\vee \neg q##. This may be specified as a replacement rule, or as the definition of one or the other of ##\to## or ##\vee##.

Using that replacement, are you able to do the problem using DNE (##\neg\neg q\equiv q##) and then DS?
 
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andrewkirk said:
These problems always depend on what rules of inference, axioms and replacement rules you've been given.
##p\to \neg q## is equivalent to ##\neg p\vee \neg q##. This may be specified as a replacement rule, or as the definition of one or the other of ##\to## or ##\vee##.

Using that replacement, are you able to do the problem using DNE (##\neg\neg q\equiv q##) and then DS?

I've got it now, thank you. I wasn't seeing how ##neg neg q \equiv q ## would fit to DS, but now it's clear.
 

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