Equivalence of logic propositions The statements:

*

*$(p\wedge q)\rightarrow r$

*$(p\rightarrow r) \vee (q\rightarrow r)$
are equivalent, as can be seen by constructing truth tables, or by applying a series of more elementary logical equivalences. I have verified this. However, I'm having trouble making sense of this in the context of an example. Namely, consider these statements about a natural number $n$:

*

*$p=$ $n$ is greater than $2$

*$q=$ $n$ is prime

*$r=$ $n$ is odd

In this case, it is true that $(p\wedge q)\rightarrow r$. "If $n$ greater than $2$ and prime, then $n$ is odd."
On the other hand, it is not the case that "($n>2$ implies $n$ odd) or ($n$ prime implies $n$ odd)". Neither of those is a true implication.
I must be interpreting statement 2 incorrectly, but I don't see how to make it right. Can anyone help me? Is this another case of the weirdness of material implication?
Thanks in advance.
 A: If you add a "for all $n$" to both statements, you get that "$\forall n,(p(n)\land q(n))\to r(n)$" is equivalent to "$\forall n,((p(n)\to r(n))\lor(q(n)\to r(n))$". Your example shows that this is not equivalent to "$(\forall n, p(n)\to r(n))\lor (\forall n,q(n)\to r(n))$".
If you instead mean to talk about a particular fixed $n$, then the "$(p\to r)\lor(q\to r)$" in your example is indeed true, as the other answers have explained.
A: Fix a natural number $n$, then "($n>2$ implies  odd) or ( prime implies  odd)" is most definitely true, why? well suppose $n\le2$ then the statement is obviously true since ($n>2$ implies  odd) is true, now suppose $n> 2$ then the only way for the statement to be false if for the the conditions to be false, but if ($n>2$ implies  odd) is false, then $n$ must be even but that means that $n$ is prime is false(cause we assumed $n>2$) and so ( prime implies  odd) is true.
A: let $p=q=r=n$.
either $n>2 \land n (mod 2)=0$ or $n \in P \land n (mod 2)=1$.
if $n>2$ then the second formula is true.
