# Rules of inference - Is my application of simplification in this proof, correct?

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 $$R$$.

• Your concern is right; Simplification acts on a conjunction. – Mauro ALLEGRANZA Jan 7 '18 at 12:28
• @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. – CalciumTablet Jan 7 '18 at 12:35
• 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. – Mauro ALLEGRANZA Jan 7 '18 at 12:44
• Alternatively, you have first to apply distributivy to 6 to get (R ∨ S). – Mauro ALLEGRANZA Jan 7 '18 at 12:55
• @MauroALLEGRANZA - You should put your comments in an answer. – 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