I'm trying to solve a problem of propositional logic. The problem is this:

$(H\vee P \vee L) \wedge (¬H \Rightarrow ¬P \vee ¬L ) \wedge (¬L\Rightarrow ¬P ) \wedge ¬H \Rightarrow L$

So I'm solving this problem through a conditional proof where I assume the antecendent and prove the consequent. In order to do this I proposed a lemma:

Lemma 1: $¬P$

$¬H \Rightarrow ¬P \vee ¬L$

$=\langle Def \Rightarrow \rangle $

$¬¬H \vee ¬P \vee ¬L$

$=\langle Double ¬\rangle $

$H \vee ¬P \vee ¬L$

$=\langle \text{Assumption: } ¬L\Rightarrow ¬P \rangle $

$H \vee ¬P \vee ¬P$

$=\langle \text{Assumption: } ¬H \text{ is true then } H \text{ is false }\rangle $

$false \vee ¬P \vee ¬P$

$=\langle \text{Identity }\vee\text{ and Idempotency}\rangle $


My first question is: Is valid to apply this assumption in the way that I did: $\langle\text{ Assumption: }¬L\Rightarrow ¬P \rangle $ and just replace ¬L by ¬P? I know Modus Ponens but don't know if it's valid in this case. If it is valid, can I always do this? If it's not, why?

My second question is: Now that I proved ¬P can I use it in another lemma by replacing it by true? Lemma 1 was enough to proof that ¬P is true?


Lemma 2: $L$

$H\vee P \vee L$

$=\langle \text{Lemma 1 : }¬P\equiv true, P\equiv false \rangle $

$H\vee false \vee L$

I want to know if this steps that I'm doing are valid or if there is another way to do them in a formal way.

(I don't know if this is relevant, but I'm studying with the book A Logical Approach to Discrete Math by David Gries and Fred Schneider).

Thank you.

  • $\begingroup$ To make the text in the math environment more readable, use \text{} to enclose it. If the math on either side comes too near, use \; to make whitespace. $\endgroup$ – Bobson Dugnutt Feb 21 '16 at 23:01
  • $\begingroup$ Thank you! I will edit it. $\endgroup$ – user3424545 Feb 21 '16 at 23:02

So, are you taking that first formula (the big conjunction) as hipotesis? In the affirmative case it is ok what you did when you replaced $\lnot L$ by $\lnot P$ in the disjunction but I can't see how from $H \lor \lnot P \lor \lnot P$ whic is equivalent to $H \lor \lnot P$, you conclude $\lnot P$.

For the second lemma, if you have already proved $\lnot P$ and you assume $H \lor P \lor L$ then it follows $H \lor L$

  • $\begingroup$ Yes, the big conjuntion is my hipotesis. Ow, sorry, I missed a part of my lemma. I'm going to edit it. $\endgroup$ – user3424545 Feb 21 '16 at 23:24
  • $\begingroup$ Sorry to ask, but I can always replace ¬L by ¬P if my hipotesis contains $¬L \Rightarrow ¬P$? $\endgroup$ – user3424545 Feb 21 '16 at 23:29
  • $\begingroup$ Ok, but now your conclution is valid only under the assumption that $H$ is false, so if you want that to be a valid step you must add $\lnot H$ to your premises. To what you are asking: yes, you cand do that, it is always valid. $\endgroup$ – la flaca Feb 21 '16 at 23:34

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