# Sentential Derivation (logic) help please

#### swtdelicaterose

I really can't figure this derivation out.

Assumptions:
1. $$\displaystyle \neg (F \vee G)$$ iff $$\displaystyle (\neg F \supset \neg F)$$
2. $$\displaystyle \neg G \supset F$$

I need to derive contradicting sentences to show that it is inconsistent, I can tell it is since $$\displaystyle \negF\supset\negF$$ is always true, which means ~(F v G) is true, so F and G both have to be false individually, which would contradict the $$\displaystyle \neg G \supset F$$ since ~G would be true and F is false.

However, I don't know how to formulate this in SD (going by The Logic Book 5th edition by Bergmann). Can anyone help?

#### Ackbeet

MHF Hall of Honor
So, you're asked to produce a contradiction out of the following assumptions:

1. $$\displaystyle \neg(F\vee G)\iff(\neg F\to\neg F)$$
2. $$\displaystyle \neg G\to F.$$

Forgive me for changing notation, but I hate the horseshoe for implication. I can never remember which way it's supposed to go!

I would go about it this way:

1. First, assume $$\displaystyle F$$.
2. Then, show that $$\displaystyle \neg F\to\neg F$$.
3. Second, assume $$\displaystyle \neg F$$.
4. Then, show that $$\displaystyle \neg F\to\neg F$$.
5. The law of the excluded middle (which in natural deduction you can derive from scratch!) tells you that $$\displaystyle F\vee\neg F$$.
6. Therefore, you can conclude that $$\displaystyle \neg F\to \neg F$$.

Alternatively, you can go like this:

1. $$\displaystyle F\vee\neg F$$. Law of excluded middle.
2. $$\displaystyle \neg F\to \neg F$$ by equivalence with step 1.

I'm just following the outline of the (valid) proof you've already provided. I think once you get started, you'll be able to continue.

So, how would you write this up so far in a two-column proof? And how would you continue?

#### PiperAlpha167

I really can't figure this derivation out.

Assumptions:
1. $$\displaystyle \neg (F \vee G)$$ iff $$\displaystyle (\neg F \supset \neg F)$$
2. $$\displaystyle \neg G \supset F$$

I need to derive contradicting sentences to show that it is inconsistent, I can tell it is since $$\displaystyle \negF\supset\negF$$ is always true, which means ~(F v G) is true, so F and G both have to be false individually, which would contradict the $$\displaystyle \neg G \supset F$$ since ~G would be true and F is false.

However, I don't know how to formulate this in SD (going by The Logic Book 5th edition by Bergmann). Can anyone help?

If you have her rules of replacement to play with, then a formal derivation of a self-contradiction is pretty straightforward.
On the other hand, if by SD you mean precisely SD and nothing more,
then (it still may be considered straight forward, but) it's going to be lengthy.

Let's suppose you have the replacement rules; then the self-contradiction (G V F) & ~(G V F) can be reach in ten lines.
At that point you're done.

All you need to reach that particular contradiction are the following rules applied in "some order":

Primitive rules of inference: R, ->I, <->E, &I
Rules of replacement: Com, Impl, DN

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