Proof

We want to prove that the mixing-up the sufficient and necessary conditions is a fallacy. * The fallacy can be formulated in propositional logic as $(p\Rightarrow q)\wedge \neg p\Rightarrow \neg q.$ * The definitions of the implication "$\Rightarrow$" and the negation " $\neg$ " give us the following truth table of the function:

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

In the second row, the premises $p\Rightarrow q$ and $\neg p$ are both true, while the conclusion $\neg q$ is false.
This is, by definition, the criterion for a fallacy.
∎

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

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

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