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Assume that in a formal proof I have

$T \cup \{ \varphi \} \vdash \varphi$

$T \cup \{ \varphi \} \vdash \lnot \varphi$

Question 1: can I then deduce $T \cup \{ \varphi \} \vdash \lnot \varphi \land \varphi$? I think there should be a rule of deduction that tells me that I can do that but there is no such rule in my lecture notes. What I do have is the following:

$\{ \psi , \varphi \} \vdash \psi \land \varphi$

So I guess my question boils down to the following:

If I have $T \vdash \varphi$, can I do $T \cup \{ \varphi \}\vdash $?

Question 2: is similar. If I have

$T \cup \{ \varphi \} \vdash \varphi \land \lnot \varphi$ ,can I deduce $T \cup \{ \varphi \} \vdash \lnot \varphi$?

Many thanks for your help.

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You should provide us with your axioms and deduction rules since there exist many deduction systems. If your axioms only have $\to$ remember that $\lnot\phi\land\phi$ is an abbreviation of $\lnot(\lnot\phi\to\lnot\phi)$. – Apostolos Oct 24 '11 at 8:36
up vote 4 down vote accepted

Firstly, $\{\varphi\}\vdash\varphi$ regardless to anything. Simply because there is a proof "$\varphi$".

If you have $T\cup\{\varphi\}\vdash\lnot\varphi$ then you have a contradiction. Simply you can write a proof for $\lnot\varphi$, write $\varphi$ and use the fact that if $\alpha$ and $\beta$ appear in your proof then you can write $\alpha\land\beta$.

From the same idea you can have that if $T\vdash\varphi$, and $T\cup\{\varphi\}\vdash\psi$ then $T\vdash\psi$. Simply by writing the proof for $\varphi$ and then using it later on in your proof of $\psi$.

In the second question you have that $T\cup\{\varphi\}$ proves a contradiction. Using explosion principle you can prove anything you want from it, $\lnot\varphi$ included (and by this xkcd strip even certain phone numbers...)

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Thanks! And what about $T \vdash \varphi$ and then $T \cup \varphi \vdash$? – Rudy the Reindeer Oct 24 '11 at 8:48
@Matt, if $T\vdash\phi$ then $T\cup\phi$ proves exactly what $T$ proves. – Apostolos Oct 24 '11 at 8:50
@Apostolos: Thanks, I just added that to my answer as well :-) – Asaf Karagila Oct 24 '11 at 8:52

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