-2
$\begingroup$

Suppose formal system $X$ has the rule $\vdash\beta\Rightarrow\hspace{2pt}\vdash\gamma$. Is the rule $\vdash\alpha\wedge\beta\Rightarrow\hspace{2pt}\vdash\alpha\wedge\gamma$ (usually) just admissible and not derivable? How may we derive the rule $\vdash\alpha\wedge\beta\Rightarrow\hspace{2pt}\vdash\alpha\wedge\gamma$ if it is derivable?

Edit:

I now see that what I needed was $\vdash(\alpha\vee\lnot\gamma)\Rightarrow\hspace{2pt}\vdash(\alpha\vee\lnot\beta)$ on the condition that $\vdash\beta\Rightarrow\hspace{2pt}\vdash\gamma$. That would up to a point have been a meaningful question as opposed to the one I posed.I will delete this question later today.

$\endgroup$

closed as off-topic by Frode Bjørdal, Todd Trimble Oct 25 '15 at 22:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Frode Bjørdal, Todd Trimble
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ (Closed on request.) $\endgroup$ – Todd Trimble Oct 25 '15 at 22:40
1
$\begingroup$

It is derivable using the Conjunction Elimination and Introduction rules.

$\endgroup$
  • $\begingroup$ Thanks Bjørn! I now have to think about whether this is really what I need to pull things together in a somewhat unusual context. Perhaps I posed the wrong question. As it is Saturday evening and I am with others, I'll postpone thinking more about this until tomorrow. $\endgroup$ – Frode Bjørdal Oct 24 '15 at 23:24
  • $\begingroup$ Kjos-Hansen I now see that what I needed was $\vdash(\alpha\vee\lnot\gamma)\Rightarrow\hspace{2pt}\vdash(\alpha\vee\lnot\beta)$ on the condition that $\vdash\beta\Rightarrow\hspace{2pt}\vdash\gamma$. That would up to a point have been a meaningful question as opposed to the one I posed. $\endgroup$ – Frode Bjørdal Oct 25 '15 at 13:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.