# Method of Truth Tables/Proof of Logical Implication

## Proof Technique

Suppose we wish to prove that $P \vdash Q$ for two propositional formulas $P$ and $Q$.

- Example: Let $P$ be $\neg p$ and $Q$ be $p \implies q$.

### Express two statements as a single WFF

We express $P \vdash Q$ as a single WFF $\left({P \implies Q}\right)$ and perform the method of truth tables on that

Demonstrating this with the example given:

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

As can be seen, the column under the main connective $\implies$ of $\left({P \implies Q}\right)$ is all $T$, so $\left({\left({\neg p}\right) \implies \left({p \implies q}\right)}\right)$ is a tautology.

Hence from Equivalence of Logical Implication and Conditional, $\neg p \vdash p \implies q$.

$\blacksquare$

### Compare two WFFs in the same table

Alternatively, we can place the two WFFs side by side in the same truth table:

$\begin{array}{cc||cc||ccc} p & q & \neg & p & p & \implies & q \\ \hline F & F & T & F & F & T & F \\ F & T & T & F & F & T & T \\ T & F & F & T & T & F & F \\ T & T & F & T & T & T & T \\ \end{array}$

This time, we need to make sure that when the truth values in the columns under the first main connectives is $T$, then it is also $T$ under the second.

Note that this is exactly the same as putting a $\implies$ between the two and making a WFF out of the pair of them.

$\blacksquare$