MHB A question on consistency in propositional logic.

AI Thread Summary
The discussion centers on a theorem in natural deduction regarding the inconsistency of hypotheses. It highlights a specific case where the set H contains a single atomic proposition p0, leading to the conclusion that {p0} implies ~p0, which is deemed impossible. Participants clarify that the theorem requires negation of phi, and in this case, phi should be p0, resulting in {p0} implying p0 instead. This correction emphasizes the importance of correctly applying the theorem's conditions. The conversation concludes with an acknowledgment of the misunderstanding in the initial application of the theorem.
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Hi everybody!

We have a theorem in natural deduction as follows:
Let H be a set of hypotheses:
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H U {~phi) is inconsistent => H implies (phi).
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Now the question arises:

Let H={p0} for an atom p0. So H U{~p0}={p0 , ~p0}.
We know that {p0 , ~p0} is inconsistent, so by our theorem we would have:
{p0} implies ~p0.
Which we know is impossible.(because for example it means that ~p0 is a semantical consequence of p0).

Now what's wrong here?
Thanks
 
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Well, your theorem or schema is negating the phi, which you're not doing. In your example, you should end up with {p0} implies p0. No doubt Evgeny can correct any mistakes I just made.
 
Ackbach said:
Well, your theorem or schema is negating the phi, which you're not doing. In your example, you should end up with {p0} implies p0.
You are right. If we apply the theorem to H U{~p0}, then phi from the theorem is p0. Therefore, the theorem concludes that {p0} implies p0.
 
Oooooooops!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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