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Proposition: If $x$ is even and $x$ is greater than $2$, then there exist prime numbers $p$ and $q$ such that $x = p + q$.

Contrapositive: If for all prime numbers $p$ or $q$, $x$ does not equal $p + q$, then $x$ is odd or $x$ is less than or equal to $2$.

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Yes, your contrapositive is pretty much correct. All I would do is change the word "or" to "and":

Contrapositive: If for all prime numbers $p$ $\color{red}{\text{and}}$ $q$ we have that $x \neq p + q$, then $x$ is odd or $x \leq 2$.

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  • $\begingroup$ how come the or doesn't get negated? $\endgroup$ – user143251 Apr 15 '14 at 9:14
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    $\begingroup$ When we say: "there exist prime numbers $p$ and $q$ such that...", it is equivalent to saying: "there exists a prime number $p$ and there exists a prime number $q$ such that...". So we're not really applying DeMorgan's Law; we're just flipping two existential quantifiers into universal quantifiers. $\endgroup$ – Adriano Apr 15 '14 at 9:17

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