1. ## Consequence relation

Hi guys, I just want to run this proof by you, just to see if I'm correct. I have to prove the following:

$\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$

where,

$\displaystyle Cn(A) = \{\beta : A \vdash \beta \}$

Firstly, given that $\displaystyle A \subseteq \text{Cn}(A)$, and that if $\displaystyle A \subseteq B$, then $\displaystyle \text{Cn}(A) \subseteq \text{Cn}(B)$, we have that $\displaystyle \text{Cn}(A) \subseteq \text{Cn}(\text{Cn}(A))$.

Secondly, pick a $\displaystyle \gamma \in \text{Cn}(\text{Cn}(A))$. Then this is such that for some $\displaystyle \beta$, with $\displaystyle A \vdash \beta$, we have that $\displaystyle \beta \vdash \gamma$. But, given the transitivity property, $\displaystyle A \vdash \gamma$, to conclude that $\displaystyle \gamma \in \text{Cn}(A)$. Thus, $\displaystyle \text{Cn}(\text{Cn}(A)) \subseteq \text{Cn}(A)$.

Combining the two, we conclude that $\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$.

HTale

2. Originally Posted by HTale
$\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$ where, $\displaystyle Cn(A) = \{\beta : A \vdash \beta \}$
Secondly, pick a $\displaystyle \gamma \in \text{Cn}(\text{Cn}(A))$. Then this is such that for some $\displaystyle \beta$, with $\displaystyle A \vdash \beta$, we have that $\displaystyle \beta \vdash \gamma$. But, given the transitivity property, $\displaystyle A \vdash \gamma$, to conclude that $\displaystyle \gamma \in \text{Cn}(A)$. Thus, $\displaystyle \text{Cn}(\text{Cn}(A)) \subseteq \text{Cn}(A)$.
Combining the two, we conclude that $\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$.
That works. But just as a matter of style, I would change the order somewhat.

$\displaystyle \begin{array}{l} \gamma \in {\rm{Cn}}\left( {{\rm{Cn(A)}}} \right)\, \Rightarrow \,\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right] \\ \beta \in Cn(A)\, \Rightarrow \,A \vdash \beta \\ A \vdash \beta \wedge \beta \vdash \gamma \, \Rightarrow \,A \vdash \gamma \\ \end{array}$

3. Originally Posted by Plato
That works. But just as a matter of style, I would change the order somewhat.

$\displaystyle \begin{array}{l} \gamma \in {\rm{Cn}}\left( {{\rm{Cn(A)}}} \right)\, \Rightarrow \,\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right] \\ \beta \in Cn(A)\, \Rightarrow \,A \vdash \beta \\ A \vdash \beta \wedge \beta \vdash \gamma \, \Rightarrow \,A \vdash \gamma \\ \end{array}$
That looks a whole lot more elegant than what I've written! The only problem is, this was a problem set today, in our first ever lecture on logic, so I'm new to some of the notation that you've presented above. Do you mind explaining what the square and round brackets mean in the first line?

4. The only notation I used that you did not is $\displaystyle P\, \Rightarrow Q$.
It reads "If P then Q". It means that if P is true then Q must also be true.

5. Originally Posted by Plato
The only notation I used that you did not is $\displaystyle P\, \Rightarrow Q$.
It reads "If P then Q". It means that if P is true then Q must also be true.
Sorry, what I meant was this notation: $\displaystyle \left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$. Why the square brackets, and why the rounded brackets? Is there any significance?

6. Originally Posted by HTale
Sorry, what I meant was this notation: $\displaystyle \left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$. Why the square brackets, and why the rounded brackets? Is there any significance?
$\displaystyle \left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$
It reads "There is a beta in Cn(A) such that beta yields gamma."

7. Originally Posted by Plato
$\displaystyle \left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$
It reads "There is a beta in Cn(A) such that beta yields gamma."
Thanks for that! I'll use that notation instead