(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:
$$\begin{array}{rl}
(p\vee q)\wedge (\neg p)&\\
(p \wedge(\neg p))\vee (q\wedge (\neg p))&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\
0\vee (q\wedge (\neg p))&(\neg p\text{ is the complement element of }p)\\
q\wedge (\neg p)&(0\text{ is the smallest element of the$B$```
})\\
\end{array}$$
* 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.
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```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()
∎

### References

#### Bibliography

**Kane, Jonathan**: "Writing Proofs in Analysis", Springer, 2016

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