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This is a simple propositional logic demonstration. I’d appreciate your help. I don’t know if my answer is correct, but the textbook used another demonstration.

The question

  1. $T \vee R$
  2. $(T \vee R) \vee (S.P) \to (Q.S)$, therefore $Q.S$

My answer:

  1. $T$ 1, Simplification
  2. $T \vee (S.P) \to (Q.S)$ 2, Simplification
  3. $T \to Q.S$ 2,4, Simplification
  4. $T$ 5, Modus Ponens

I know this is a very basic question and my answer is not elegant, but since I’m studying logic all by myself, and have no teacher to ask, I’d appreciate an answer.

The book made a more elegant and simple answer:

  1. $(T \vee R) \vee (S.P)$ 1, Addition
  2. $Q.S$ 2,3 Modus Ponens
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You cannot apply Simplification to a Disjunction.

$X\vee Y$ does not logically entail $X$.

$X\vee Y$ means at least one from $\{X, Y\}$ is true, which does not guarantee that one will be $X$.


You also cannot apply the rule of Simplification to the antecedent of a Conditional statement.


The rule of Simplification is applied to a Conjunction, when that connective is the root operation in the statement.$$X\wedge Y~\vdash~X\\X\wedge Y~\vdash~Y$$

$X\land Y$ means both from $\{X,Y\}$ are true, and that entails that $X$ is true.

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  • $\begingroup$ Thank you! It was a simple and elementary mistake indeed! 🤦🏻‍♀️ $\endgroup$ – brigittethecat Apr 19 at 12:55

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