Could someone verify that my proof is valid as the question did not have a solution?

  1. $(\lnot R \lor \lnot F) \to (S \land L)$
  2. $(S \to T)$
  3. $(\lnot T)$
  4. $(\lnot S)$ 2,3 Modus tollens
  5. $\lnot (\lnot R \lor \lnot F) \lor (S \land L)$ 1, implication equivalence
  6. $(R \land F) \lor (S \land L)$ 5, double negation
  7. $(R \lor S)$ 6, simplification
  8. $(R)$ 7,4 Disjunctive syllogism

My main concern is with line 7 with the use of simplification, have I applied the rule correctly?
I understand that with simplification if you have $(P \land Q)$ and apply it, it returns $(P)$.

For my proof, you had to show that lines 1-3 (the hypotheses) entail $R$.

  • 1
    $\begingroup$ Your concern is right; Simplification acts on a conjunction. $\endgroup$ – Mauro ALLEGRANZA Jan 7 '18 at 12:28
  • $\begingroup$ @Maruo ALLEGRANZA So is what I've done incorrect, as I've applied simplification to line 6 however I only did so for R and S removing F and L from it. $\endgroup$ – CalciumTablet Jan 7 '18 at 12:35
  • 2
    $\begingroup$ instead of implic equiv in 5, you can assume ¬R and use addition to get : (¬R ∨ ¬F). With it, by MP: ( S ∧ L) that - by simpl - gives S. Now you have a contradiction with 4 and conclude with ¬ ¬R and then R by double negation. $\endgroup$ – Mauro ALLEGRANZA Jan 7 '18 at 12:44
  • 1
    $\begingroup$ Alternatively, you have first to apply distributivy to 6 to get (R ∨ S). $\endgroup$ – Mauro ALLEGRANZA Jan 7 '18 at 12:55
  • $\begingroup$ @MauroALLEGRANZA - You should put your comments in an answer. $\endgroup$ – Taroccoesbrocco Jan 8 '18 at 18:17

As correctly said by Mauro Allegranza, your usage of simplification is wrong.

As an alternative to the proofs suggested by Mauro Allegranza (which are perfect), consider the following proof:

  1. $(\lnot R \lor \lnot F) \to (S \land L)$ assumption
  2. $(S \to T)$ assumption
  3. $(\lnot T)$ assumption
  4. $(\lnot S)$ 2,3 Modus tollens
  5. $(\lnot S \lor \lnot L)$ 4, addition
  6. $\lnot(S \land L)$ 5, De Morgan
  7. $\lnot(\lnot R \lor \lnot F)$ 1, 6 Modus tollens
  8. $(R \land F)$ 7, double negation (De Morgan)
  9. $(R)$ 8, simplification
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