Sum of any two prime numbers except the prime number 2 is even.

Question

(a) For every odd natural number, there is a different natural number such that their sum is even.

$$(\forall x): (\neg \text{Even}(x) \rightarrow (\forall y): (y \neq x \rightarrow \text{Even}(x+y)).$$

Alternate Solution: $\forall x (\neg \text{Even}(x) \rightarrow \exists y (y \neq x \land \text{Even}(x+y))$

(b) The sum of any two prime number except the prime number 2 is even.

1. $(\forall x, y):((x \neq 2 \land y \neq 2) \rightarrow ([\text{Prime}(x) \land \text{Prime}(y)) \rightarrow \text{Even}(x+y))).$

2. $(\forall x, y):((\text{Prime}(x) \land \text{Prime}(y) \land \text{Even}(x+y)) \rightarrow (x \neq 2 \land y \neq 2)).$

I am not sure which one of the logical statements is correct for (b), I think it's 2. because the first statement will be vacuously true even if $x$ and $y$ are primes not equal to $2$.

Is my reasoning correct? Please correct me if I am wrong. Thank you!

Answer 1

For a) consider for example x = 3. Is it true that for all natural y, x+y is even?

For b2) consider x, y = 2. Would then statement be true?

For b1) what happens if x = 7 and y = 2?

UPD: However, all of these maybe just a minute of dumbness from me.

Answer 2

Here is a supplementary answer on $(a)$. This isn't what you originally asked, but there are errors here too.

First, I recap the comments. Your original solution was $$ \forall x(\neg E(x)\rightarrow \forall y (y\neq x\rightarrow E(x+y))) $$ (I'm using $E$ for the Even predicate).

So right away one can tell this is probably wrong because the quantifier on $y$ doesn't match up. The original sentence says there is a different natural number, while the formula has $\forall y$.

However, there is more to fixing this than simply changing $\forall$ to $\exists$, because these two quantifiers act differently when attached to implications. Consider the following two statements:

1. $x+y$ is even for every $y$ different from $x$.
2. $x+y$ is even for some $y$ different from $x$.

The first attempt expresses option 1 with the clause "$\forall y(y\neq x\rightarrow E(x+y)$" or, in other words, "for all $y$, if $y\neq x$ then $x+y$ is even". So this accurately describes (1). But, of course, you don't want (1), you want (2). So what happens if I just change $\forall$ to $\exists$? I get:

$$ \exists y (y\neq x\rightarrow E(x+y)) $$

So does this accurately express (2)? Translating directly, I get "there is a $y$ such that if $y\neq x$ then $x+y$ is even". This does not have the same meaning as "$x+y$ is even for some $y\neq x$". So one needs to think about how to accurately express "there is a $y$ different from $x$ such that $x+y$ is even".