Proof using Rule of Disjunctive Amplification

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Discussion Overview

The discussion revolves around the application of the rule of disjunctive amplification in logical proofs, specifically how it relates to negated propositions. Participants explore the implications of this rule and whether it can be applied to negated statements in the same way as positive statements.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions how the conclusion ##\neg r \lor \neg s## is derived from ##\neg r## using disjunctive amplification, noting that the rule is typically stated as ##p \implies p \lor q##.
  • Another participant suggests that the proof can be simplified by letting ##p = \neg r## and ##q = \neg s##, implying that this substitution is valid.
  • A later reply challenges the idea that negating propositions allows for the same logical rules to apply, questioning whether rules like Modus Ponens can also be transformed in this way.
  • One participant asserts that ##p## can represent any proposition, including negated ones, and mentions the law of substitution as relevant to this discussion.
  • Another participant emphasizes the validity of deriving ##\neg q## from ##\neg p## and ##\neg p \implies \neg q##, reinforcing the connection to Modus Ponens.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of logical rules to negated propositions, with some agreeing on the validity of substitutions while others raise concerns about the implications of such transformations. The discussion remains unresolved regarding the broader applicability of these transformations across different logical rules.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the applicability of logical rules to negated propositions, and the potential scope of these transformations is not fully explored.

hotvette
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TL;DR
use of negation in disjunctive amplification
Book shows a proof where a conclusion is reached of: ##\neg r##. The next step says ##\neg r \lor \neg s## using the rule of disjunctive amplification. The rule of disjunctive amplification as I know it is ##p \implies p \lor q##. I don't see how from this you can also say ##\neg p \implies \neg p \lor \neg q##. I can see that the truth table is a tautology so I know it's true, I just don't see how to get there.
 
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Is this as simple as letting ##p = \neg r## and ##q = \neg s##?
 
hotvette said:
Is this as simple as letting ##p = \neg r## and ##q = \neg s##?
Yes.
 
Really? So this means I can take any of the logic Rules and just replace anything with its negative and vice versa and it is still valid? Example of M. Ponens ##[p \land (p \implies q] \implies q## can be written as ##[\neg p \land (\neg p \implies \neg q] \implies \neg q##? The book makes no mention of this. I wonder how we are expected to know...
 
##p## is any proposition, including ##\neg q## (kind of. Technically it follows the law of substitution, but functionally it’s the same thing).
 
Think through it carefully. If you have ##\neg p## and ##\neg p \implies \neg q##, why wouldn’t you have ##\neg q##?
 
Sure, that's just M. Ponens. Makes sense, thanks!
 

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