A question on consistency in propositional logic.

[Mathelogician]

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

 

[Ackbach]

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.

 

[Evgeny.Makarov]

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.

 

[Mathelogician]

Oooooooops!

 

[blue_raver22]

for your question!

In this scenario, the inconsistency arises because the set of hypotheses includes both p0 and its negation, ~p0. This creates a logical contradiction, which means that any statement can be derived from this set of hypotheses. Therefore, when we apply the theorem, we get the result that p0 implies ~p0, which is impossible because it would mean that both p0 and ~p0 are true at the same time.

This highlights the importance of consistency in propositional logic. In order for a set of hypotheses to be valid, they must not contain any contradictions. Otherwise, any conclusion drawn from that set would be unreliable. In this case, the inconsistency arises from a simple mistake of including both p0 and ~p0 in the set of hypotheses. It is important to carefully construct and evaluate sets of hypotheses in order to ensure consistency and validity in propositional logic.