Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am struggling with the way to write a clear and mathematical proof of logical theorems. Take for example the theorem $\Gamma \models A, \Gamma \subseteq \Delta$ implies $\Delta \models A$. I can prove it this way:

If every formula in $\Delta$ is validated, then that validates every formula in $\Gamma$, which validates $A$.

But how do I prove such thing with mathematical language?

share|improve this question
    
What exactly dies "validated" mean here? Spell that out a bit, and you will have your desired proof. –  Peter Smith May 15 '13 at 18:19
1  
If the word "validated" was used in the definition of $\models$ (and was, therefore, defined beforehand), then you might not need to spell it out, but otherwise I agree with Peter that you should. –  Andreas Blass May 15 '13 at 20:55

1 Answer 1

A proof of the above fact needs the definitions :

(i) $Γ \vDash A$ iff for every interpretation $\mathcal M$, if all formulae in $\Gamma$ are true in $\mathcal M$, then also $A$ is

and :

(ii) $\Gamma \subseteq \Delta$ iff for every formula $\alpha$, if $\alpha \in \Gamma$, then $\alpha \in \Delta$.

Consider now an interpretation $\mathcal M$: due to the fact that $\Gamma \subseteq \Delta$, if all formulae in $\Delta$ are true in $\mathcal M$, then $\mathcal M$ will be also a model for all formulae in $\Gamma$ (which are among those in $\Delta$).

But $A$ is true in every model of $\Gamma$; thus, for every interpretation $\mathcal M$, if $\mathcal M$ is a model for all formulae in $\Delta$, then also $A$ is true in $\mathcal M$, i.e.

$\Delta \vDash A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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