# Proof

(related to Lemma: Disjunctive Syllogism)

We want to prove that the disjunctive syllogism is a valid logical argument. * The disjunctive syllogism can be formulated in propositional logic as $((p\vee q)\wedge (\neg p))\Rightarrow q.$ * On the left side of the disjunctive syllogism we have only the operations "$\wedge$" and "$\vee$", we can use the fact that propositional logic is a Boolean algebra $(B,\wedge,\vee,1,0)$. Applying the properties of this Boolean algebra $B$ we can conclude that: $$\begin{array}{rl} (p\vee q)\wedge (\neg p)&\\ (p \wedge(\neg p))\vee (q\wedge (\neg p))&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\ 0\vee (q\wedge (\neg p))&(\neg p\text{ is the complement element of }p)\\ q\wedge (\neg p)&(0\text{ is the smallest element of theB})\\ \end{array}$$ * Therefore, by definition of a valid logical argument, the disjunctive syllogism is valid if and only if $q$ is true, whenever both $q$ and $\neg p$ are true. But this is trivial.

 This result can also be achieved by calculating the truth table of the whole expression of the disjunctive syllogism, which confirmes that it is a tautology. Please click evaluate to verify this: import sage.logic.propcalc as propcalc f = propcalc.formula("((p|q)&(~p))->q") f.truthtable() ∎ Thank you to the contributors under CC BY-SA 4.0!  Github: References Bibliography Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016 
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