3. Logical consequence

Logical consequence

Argument

An argument is a sequence of [[Proposition|propositions]].

$$\begin{array}{rlrl}& \underline{\begin{array}{l}{p}_1\\ p_2\\ ..\\ p_n\end{array}\}}& & n\text{ premises}\\ \therefore & q& & \text{one conclusion}\end{array}$$

You can read ”$p_1,p_2,...,p_n$, therefore $q$”.

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Logical consequence

In a valid argument, we can informally say that conclusion $B$ “follows” from a set of premises $A_1,...,A_n$.

Formally, we say that formula $B$ is a logical consequence of the set of formulae $\{A_1,...,A_n\}$, if the following implication holds for every interpretation $v$: $$ \text{If } v(A_i) = 1, \text{ for all } 1 \le i \le n, \text{ then } v(B) = 1. $$

Alternative terminologies include:

• $B$ is a logical consequence of $A_1,...,A_n$.
• $A_1,...,A_n\models B$.
• The argument $A_1,...,A_n\models B$ is valid.
• $B$ is semantically entailed (or implied) by $A_1,...,A_n$
• $B$ is a valid consequence of $A_1,...,A_n$

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Alternate definition of Model

Let the symbol $\mathcal{I}$ denote the set of all interpretations. Let $\mathcal{S}=\{A_1,...,A_n\}$. We have that

$$\mathcal{S}\models B⟺\mathrm{mod}(\mathcal{S})\subseteq \mathrm{mod}(B)$$

Notice that $\mathrm{mod}(\neg B)=\mathcal{I}-\mathrm{mod}(B)$ hence:

$$\mathrm{mod}(B)\cap \mathrm{mod}(\neg B)=\varnothing$$

Therefore, if $\mathcal{S}\models B$, then $\mathrm{mod}(\mathcal{S})\cap \mathrm{mod}(\neg B)=\varnothing$.

title: Example
Show that $P, P \rightarrow Q \models Q$.
 
| *interpretation* | $p$ | $q$ | $p \rightarrow q$ | $p \land (p \rightarrow q)$ | $q$ |              $\models$              |
|:----------------:|:---:|:---:|:-----------------:|:---------------------------:|:---:|:-----------------------------------:|
|     $v_{0}$      |  0  |  0  |         1         |              0              |  0  |        ✅ **premises false**        | 
|     $v_{1}$      |  0  |  1  |         1         |              0              |  1  |        ✅ **premises false**        |
|     $v_{2}$      |  1  |  0  |         0         |              0              |  0  |        ✅ **premises false**        |
|     $v_{3}$      |  1  |  1  |         1         |              1              |  1  | ✅ **premises and conclusion true** |
 
The statement follows because in every row in which the columns for $P$ and $P \rightarrow Q$ contain $1$ (the premises), so does the column for $Q$ (the conclusion).
 
$$
	\bmod (\{ P, P \rightarrow Q \}) = \{ v_3 \} \subseteq \, \bmod(\{ Q \}) = \{ v_1, v_3 \}
$$
 
This type of derivation is called [[Modus ponens]].

Special case: unsatisfiable premises

Take the following scenario:

• If Jack takes a holiday, then Jill will be happy and she will not cry.
• Jack will take a holiday and if Jill is happy she will cry.
• Therefore Jack will take a holiday.

We define this formally using:

• Let $J$ stand for ‘Jack will take a holiday’
• Let $H$ stand for ‘Jill will be happy’
• Let $C$ stand for ‘Jill will cry’

In this case, the argument $J\to (H\wedge \neg C),J\wedge (H\to C)\models J$ is valid. This is because $\mathrm{mod}(\mathcal{S})=\varnothing \subseteq \mathrm{mod}(B)$. In general, any conclusion follows from an unsatisfiable set of premises.

Special case: tautological conclusions

Tautologies are always true, so if $A$ is a tautology, then $\mathrm{mod}(A)=\mathcal{I}$.

This has two effects:

1. A tautology is a logical consequence of any set of formulae.
2. A tautology also follows from ‘nothing’. Any interpretation satisfies all of the formulae in the empty set, hence $\mathrm{mod}(\varnothing )=\mathcal{I}$. However, if $A$ is a tautology, then $\mathrm{mod}(A)=\mathcal{I}$ hence $\mathrm{mod}(\varnothing )=\mathcal{I}\subseteq \mathrm{mod}(A)=\mathcal{I}$.

Invalid argument

Invalid argument

An argument that is not valid is said to be invalid.

By the definition of logical consequence, $A_1,...,A_n/\models B$ if there exists an interpretation $v$ such that:

$$v(A_i)=1\text{ for all }1\le i\le n,\text{ but }v(B)=0$$

Hence, in order to show a conclusion does not follow from a set of premises, we must find an interpretation that makes all of the premises true, but under which the conclusion is false.

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title: Example 1
Show that $P \not\models Q$ where $P$ and $Q$ are atoms.
 
**Solution**: Take the interpretation $v$ with $v(P) = 1$ and $v(Q) = 0$.

title: Example 2
Show that $P \rightarrow Q \not\models Q$ where $P$ and $Q$ are atoms.
 
**Solution**: Take the interpretation $v$ with $v(P) = 0$ and $v(Q) = 0$.
When $P$ and $Q$ are $0$, it makes $P \rightarrow Q$ true but $Q$ is still false.