Consequence relation

**HTale** · January 22nd 2009, 12:23 PM

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

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

where,

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

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

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

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

Thanks in advance,

HTale

**Plato** · January 22nd 2009, 12:42 PM

Originally Posted by HTale

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

That works. But just as a matter of style, I would change the order somewhat.

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

**HTale** · January 22nd 2009, 12:56 PM

Originally Posted by Plato

That works. But just as a matter of style, I would change the order somewhat.

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

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?

Thanks in advance.

**Plato** · January 22nd 2009, 01:14 PM

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

**HTale** · January 22nd 2009, 01:29 PM

Originally Posted by Plato

The only notation I used that you did not is $P\, \Rightarrow Q$ .
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: $\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$ . Why the square brackets, and why the rounded brackets? Is there any significance?

**Plato** · January 22nd 2009, 01:32 PM

Originally Posted by HTale

Sorry, what I meant was this notation: $\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$ . Why the square brackets, and why the rounded brackets? Is there any significance?

$\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$
It reads "There is a beta in Cn(A) such that beta yields gamma."

**HTale** · January 22nd 2009, 01:34 PM

Originally Posted by Plato

$\left( {\exists \beta \in Cn(A)} \right)\left[ {\beta \vdash \gamma } \right]$
It reads "There is a beta in Cn(A) such that beta yields gamma."

Thanks for that! I'll use that notation instead

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