(related to Lemma: Modus Ponens)
We want to prove that modus ponens is a valid logical argument.
* Modus ponens can be formulated in propositional logic as $((p\Rightarrow q)\wedge p)\Rightarrow q.$
* Reformulating the implication as disjunction gives us the expression $((\neg p\vee q)\wedge p)\Rightarrow q.$
* On the left side we have only the operations "$\wedge$" and "$\vee$". Therefore, we can use the fact that propositional logic is a Boolean algebra $(B,\wedge,\vee,1,0)$, and make use of the properties of this Boolean algebra $B$ as follows:
$$\begin{array}{rl}
((\neg p\vee q)\wedge p)\Rightarrow q&\\
(p \wedge(\neg p\vee q))\Rightarrow q&(\text{commutativity of }"\wedge")\\
((p \wedge\neg p)\vee (p\wedge q))\Rightarrow q&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\
(0\vee (p\wedge q))\Rightarrow q&(\neg p\text{ is the complement element of }p)\\
(p\wedge q)\Rightarrow q&(0\text{ is the smallest element of the$B$})\\
\end{array}$$
* It follows from the truth table of conjunction that if $p$ and $q$ are true, than $q$ is true.
* By definition of a valid logical argument, modus ponens $((p\Rightarrow q)\wedge p)\Rightarrow q$ is valid.
