Logical Consequence 2

Continuing on the same theme, consider the alphabet $A=\{0,1,2\}$. Consider the strings:

${\sigma }_1=(0,1,2)$
${\sigma }_2=(0,1,1)$.

Consider the question:

Is ${\sigma }_2$ a logical consequence of ${\sigma }_1$?

Clearly the question doesn't make sense.

Next, we turn this alphabet $A$ into a formalized language $L$. $String(L)$ is the set of finite sequences drawn from $A$. We define $Sent(L)$--i.e., the sentences of $L$--inductively as follows:

$Sent(L)$ is smallest subset $X\subseteq String(L)$ such that:

(i) $(1)\in X$;
(ii) $(2)\in X$;
(iii) if ${\sigma }_1,{\sigma }_2\in X$, then $(0)\ast {\sigma }_1\ast {\sigma }_2\in X$.

(Here $\ast$ is sequence concatenation.) So, we get:

$(1)\in Sent(L)$;
$(2)\in Sent(L)$;
$(0,1,1)\in Sent(L)$.
$(0,1,2)\in Sent(L)$.
etc.

So, these strings are sentences in $L$.
For brevity, I now write

$012$

to mean the sequence

$(0,1,2)$

So,

$1\in Sent(L)$;
$2\in Sent(L)$;
$011\in Sent(L)$.
$012\in Sent(L)$.
etc.

One can enumerate the $L$-sentences by partioning them by length:

1. There are no $L$-sentences of length $k$, where $k$ is even.
2. The $L$-sentences of length 1 are $1$ and $2$.
3. The $L$-sentences of length 3 are $011,012,021$ and $022$.
4. etc.

So far, there is just the alphabet $A$, the $L$-strings and the $L$-sentences. One has no notion of logical consequence.
Let $B_2$ be the two-element Boolean algebra, $\{\mathrm{\top },\mathrm{\perp }\}$.

An $L$-interpretation is a function $I:Sent(L)\to \{\mathrm{\top },\mathrm{\perp }\}$ such that, for any $\varphi ,\theta \in Sent(L)$,

$I(0\ast \varphi \ast \theta )=\mathrm{\top }$ iff $I(\varphi )=\mathrm{\top }$ and $I(\theta )=\mathrm{\top }$.

Write

$I\models \varphi$

for

$I(\varphi )=\mathrm{\top }$.

Let $\mathrm{\Sigma }(L)$ be the class of such interpretations.
Finally, logical consequence is defined by:

$\theta$ is a logical consequence in $L$ of $\varphi$ iff $\varphi ,\theta \in Sent(L)$ and, for all $I\in \mathrm{\Sigma }(L)$, if $I\models \varphi$, the $I\models \theta$.

We abbreviate this relationship as.

$\varphi {\models }_L\theta$

Then we have:

$1{\models }_{L}1$.
$1{⊭}_{L}2$.
$2{⊭}_{L}1$.
$2{\models }_{L}2$.
$011{\models }_{L}1$.
$011{⊭}_{L}2$.
$012{\models }_{L}1$.
$012{\models }_{L}2$.
$021{\models }_{L}1$.
$021{\models }_{L}2$.
$022{\models }_{L}2$.
$022{⊭}_{L}1$.

This all looks quite hard to follow. Let me abbreviate a bit more. Let's write "$P$" for "$1$", and "$Q$" for "$2$" and "$\wedge$" for "$0$". (Not that this makes any difference. Strings are just strings. They have no "intrinsic" meaning.) Furthermore, instead of

$0\varphi \theta$

we write

$\varphi 0\theta$.

(I.e., infix notation.)
Then the above becomes:

$P{\models }_{L}P$.
$P{⊭}_{L}Q$.
$Q{⊭}_{L}P$.
$Q{\models }_{L}Q$.
$P\wedge P{\models }_{L}P$.
$P\wedge P{⊭}_{L}Q$.
$P\wedge Q{\models }_{L}P$.
$P\wedge Q{\models }_{L}Q$.
$Q\wedge P{\models }_{L}P$.
$Q\wedge P{\models }_{L}Q$.
$Q\wedge Q{\models }_{L}Q$.
$Q\wedge Q{⊭}_{L}P$.

This all looks so much more familiar!

Now go back to the question at the start. Consider the strings:

${\sigma }_1=(0,1,2)$
${\sigma }_2=(0,1,1)$.

Consider the question:

Is ${\sigma }_2$ a logical consequence in $L$ of ${\sigma }_1$?

Clearly the question does now make sense. It asks,

Is $P\wedge P$ a logical consequence in $L$ of $P\wedge Q$?

And the answer is: yes.