# How do I interpret $F_1, \neg F_1 \vdash F_2$

I don't think I am right to say:

If I know $F_1$ and $\neg F_1$ then I know $F_2$?

Cos $F_1$ is not related to $F_2$? Or perhaps there's a typo? It says in my notes: negation elimination

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The theorem states: From a contradiction, anything follows.

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Can you give an example? –  Jiew Meng Sep 28 '12 at 13:56
Pigs can fly as soon as the hell freezes –  Hagen von Eitzen Sep 28 '12 at 13:58
@Jiew If you know both $F_1$ and $not F_1$ then you have a contradiction. A falsehood. An example is that both the propositions $1+1 = 2$ and $1+1 = 3$ are valid for whatever reason. The rule in your question then gives ANYTHING as a valid consequence. For example, you could conclude $1+1 = 73$, $pigs fly$ or whatever you like. –  Glen Wheeler Sep 28 '12 at 14:06

From a contradiction, anything follows. Here's proof:

1. $F1 \wedge \neg F1$ (Premise)

2. $\neg F2$ (Premise)

3. $F1$ (1)

4. $\neg F1$ (1)

5. $F1 \wedge \neg F1$ (3, 4)

6. $\neg \neg F2$ (Conclusion, 2)

7. $F2$ (Remove $\neg \neg$, 6)

8. $F1 \wedge \neg F1 \rightarrow F2$ (Conclusion, 1)

Alternatively, you can use a truth table to prove that $F1 \wedge \neg F1 \rightarrow F2$ is a tautology. Truth table generator at: http://mathdl.maa.org/images/upload_library/47/mcclung/index.html

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