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Proof

(related to Lemma: Mixing-up the Inclusive and Exclusive Disjunction)

We want to prove that the mixing-up the inclusive and exclusive disjunction is a fallacy. * The fallacy can be formulated in propositional logic as $(p\vee q)\wedge p\Rightarrow \neg q.$ * We get the following truth table of the function due to the definition of negation and implication:

$[[p]]_I$	$[[q]]_I$	$[[p\vee q]]_I$	$[[ \neg q]]_I$
$0$	$0$	$0$	$1$
$0$	$1$	$1$	$0$
$1$	$0$	$1$	$1$
$1$	$1$	$1$	$0$

The last row shows that while the premises $p$ and $p\vee q$ are both true, the conclusion $\neg q$ is false.
This is, by definition, the criterion for a fallacy.
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References

Bibliography

Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016
Kohar, Richard: "Basic Discrete Mathematics, Logic, Set Theory & Probability", World Scientific, 2016