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I have the following:

Premise: ((V → ¬W) ∧ (X → Y))
Premise: (¬W → Z)
Premise: (V ∧ X)
         |- (Z ∧Y)

The part I want to know is how do I go about separating the second premise ¬W so that it becomes ¬Z or (W → Z)

I have heard of doing double negation (¬¬W → ¬Z) but still unsure how to proceed from here.

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  • $\begingroup$ I can't understand what you want. The second premise is $\neg W\to Z$. What do you mean with separate? $\endgroup$ – Git Gud May 10 '15 at 14:24
  • $\begingroup$ I don't understand what you are trying to do. Can you perhaps elaborate? It sounds to me like you are trying to rewrite the premise "(¬W → Z)" to become "¬Z or (W → Z)", but this is not possible. $\endgroup$ – Mankind May 10 '15 at 14:24
  • $\begingroup$ Hi, sorry if I'm confusing you guys I am only just learning this so I may not have all the correct terminology / logic. I want to solve to (Z ∧ Y). My ultimate goal is to modify the first premise into ((V->Z) ∧ (X → Y)) but to do that I was thinking I need (¬Z)? or is there a way to easily go combine the ¬W values of premise 2 into 1? $\endgroup$ – Kree Do May 10 '15 at 14:29
  • $\begingroup$ The second premise is the one you can't split easily split up into two assumptions. But the other two have a $\land$ at the top, so they can be split. Now you have $V\to \neg W$ and $X\to Y$ and $\neg W\to Z$ and $V$ and $X$. Can you see how you can make $Z$ and $Y$ from those components? $\endgroup$ – Henning Makholm May 10 '15 at 14:30
  • $\begingroup$ @KreeDo You can't modify the first premise to include $V\to Z$, but I can see you're going in the right direction. Instead look at Makholm's hint above which is the proper way of pursuing your idea. $\endgroup$ – Git Gud May 10 '15 at 14:31
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As said in the comments, you don't need to split the second premise. Here you can use the conjunction elimination ($X\land Y \vdash X$, $X\land Y \vdash Y$), conjunction introduction ($X,Y\vdash X\land Y$) and modus ponens ($X,X\rightarrow Y\vdash Y$).

If I number you premises :

  1. $(V\rightarrow \neg W)\land (X\rightarrow Y)$
  2. $\neg W\rightarrow Z$
  3. $V\land X$

Now you can apply conjunction elimination on 1. and 3. which gives you

  1. $V\rightarrow \neg W$
  2. $X\rightarrow Y$

  3. $V$

  4. $X$

Here you can see that if you use modus ponens on 6., 4. you get $\neg W$. This $\neg W$ is on the left of premise 2. So again using modus ponens, you get $Z$.

To get $Y$, you must use modus ponens one last time with 7. and 5.

Conjunction introdution gives you $Z\land Y$.

Here is what I get in Fitch notation :

Natural deduction in Fitch notation

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  • $\begingroup$ thanks for the answer, already got it from the hits above but this is a much nicer and neater way to show it! $\endgroup$ – Kree Do May 10 '15 at 15:38

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