If $\alpha \vdash \phi$ and $\beta \vdash \phi$, does it mean that $\alpha \cap \beta \vdash \phi$?

Question

I would like to know if this is correct.

You have to specify that $\alpha, \beta$ are sets of formulas. — Mauro ALLEGRANZA, Apr 24 '18 at 11:21
@MauroALLEGRANZA Could I know why it is not correct? For example, if $\alpha$ is included in a set, and $\beta$ included in the same set, then the intersection is also included. — IliassA, Apr 24 '18 at 11:30

score 5 · Accepted Answer · answered Apr 24 '18 at 11:33

5

Consider the following example :

$A = \{ p \land q \}$ and $B = \{ q \land r \}$.

We have $A \vdash q$ and $B \vdash q$.

But $A \cap B = \emptyset$ and :

$\emptyset \nvdash q$,

because $q$ is not a tautology.

answered Apr 24 '18 at 11:33

Mauro ALLEGRANZA

But, isn't the $\emptyset$ included in all sets, which means $\emptyset \vdash $ any possible set? – IliassA Apr 24 '18 at 11:36
@IliassA No, check the meaning of $\vdash$ – Hagen von Eitzen Apr 24 '18 at 11:41
@IliassA - Yes, $\emptyset$ is included in evry set but this does not mean that every set will prove every formula. – Mauro ALLEGRANZA Apr 24 '18 at 12:03
Clear to me now, thanks. – IliassA Apr 26 '18 at 13:30

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