Method of Truth Tables/Proof of Tautology

Proof Technique
This is used to establish whether or not a given logical formula is a tautology; that, is true under all models.

Let $P$ be a logical formula we wish to validate, and write it as a WFF of Propositional Calculus.

We will use Peirce's Law as an example:
 * $\left({\left({p \implies q}\right) \implies p}\right) \implies p$

For each model of the set of atoms, we write its truth value underneath:

$\begin{array}{cc||ccccccc} p & q & ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & F & F &  & F &   & F &   & F \\ \end{array}$

In the above, we are using the model $p_{\mathcal M} = F, q_{\mathcal M} = F$.

Then we fill in the truth value of each of the WFFs on the parsing sequence of $P$, underneath its main connective:

$\begin{array}{cc||ccccccc} p & q & ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & F & F &  & F &   & F &   & F \\ &  &   & T &   &   &   &   &   \\ &  &   &   &   & F &   &   &   \\ &  &   &   &   &   &   & T &   \\ \end{array}$

In the above, this has been done on separate lines, so as to clarify the sequence in which this is done for the example.

In practice we write it like this:

$\begin{array}{cc||ccccccc} p & q & ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & F & F & T & F & F & F & T & F \\ \end{array}$

We repeat this for all models:

$\begin{array}{cc||ccccccc} p & q & ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & F & F & T & F & F & F & T & F \\ F & T & F & T & T & F & F & T & F \\ T & F & T & F & F & T & T & T & T \\ T & T & T & T & T & T & T & T & T \\ \end{array}$

In the column under the main connective of $P$ itself can be found the truth value of $P$ for each model.


 * If this contains nothing but $T$, then $P$ is a tautology.
 * If this contains nothing but $F$, then $P$ is a contradiction.
 * If this contains $T$ for some models and $F$ for others, then $P$ is a contingent statement.

In the above example, the main connective of $P$ is the rightmost instance of $\implies$.

The column beneath that connective is all $T$, so $\left({\left({p \implies q}\right) \implies p}\right) \implies p$ is a tautology.