# Proof

(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.

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### References

#### Bibliography

1. Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016