> [!NOTE] Theorem > The set of [[L2 (FOL)|L2]] connectives $\{ \land, \lor, \lnot \}$ are [[Truth Functionally Complete Set of Connectives|truth functionally complete]]: that is for any [[Truth Function|truth function]] (specified by a [[Truth Table|truth table]]), there is a sentence containing these connectives alone, that has this true table. **Proof**: Let $\varphi(P_{1},\dots P_{n})$ be a truth function specified by the following truth table: $ \begin{array}{c|c|c||l} \mathrm{P}_1 & \ldots & \mathrm{P}_{\mathrm{n}} & \varphi\left(\mathrm{P}_{1}, \ldots, \mathrm{P}_{\mathrm{n}}\right) \\ \hline \mathrm{T} & \ldots & \mathrm{T} & \\ \mathrm{T} & \ldots & \mathrm{F} & \\ \vdots & & \vdots & \\ \vdots & & \vdots & \\ \mathrm{F} & \ldots & \mathrm{F} & \end{array} $Suppose $\varphi(P_{1},\dots P_{n})$ is true at least once. For each situation in which $\varphi$ is true, we form a conjunction of the sentence letters in the following way: for $i=1,\dots,n,$ if $P_{i}$ is true, $P_{i}$ is a conjunct in this conjunction, otherwise, $\lnot P_{i}$ is a conjunct in this conjunction. Then the disjunction of all these conjunctions has the same truth table as the original truth function. On the other hand, if its truth table is false in every row, $\varphi$ can be expressed by any contradiction built from the $P_{i}$s (e.g. $P_{1}\land \lnot P_{1}$).