(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.
