What type of logical statement is this statement and how do you prove it? I'm learning proof techniques. I understand "if, then" statements and how they correspond to "if p, then q" and "hypothesis, and conclusion". But, I don't understand what kind of statement the following is considered to be:
"The sum of two even integers is always even"
How do I look at this statement logically, using p and q? Is there a hypothesis and conclusion here? Is this considered a non-conditional statement? And lastly, how I do I prove such a statement?
This was one of the first statements I came across when I started learning how to do proofs, so I have seen the proof for it. I just don't understand it logically.
Thank you for the help.
UPDATE:
Bram28 asked, "Do you have to use formal logic for this? Or would a mathematical proof suffice?"
What would formal logic look like for this proof? Or, am I misunderstanding something?
I've been reading more about ZFC Theory and http://us.metamath.org/mpegif/opoeALTV.html for instance. And, I can see how these proofs are being constructed.
That's why I am so curious as to how Bram28's idea of formal logic would apply to my original post.
 A: It's almost always helpful to write out definitions in these cases! What's an even integer? Well we say an integer $x$ is even if (and only if) $\exists k \in \mathbb{Z}\ \big(x = 2k\big)$.
So the statement is saying given any two integers $x$ and $y$ that satisfy the property above (that is the property that defines them as being even), their sum is also even. Let's denote the sum $z = x + y$. This means $\exists k \in \mathbb{Z}\ \big(z = 2k\big)$. In other words, that the statement $\exists k \in \mathbb{Z}\ \big(x+y = 2k\big)$ is true.
Since the statement is really saying that for any two integers which are even, their sum is even, we are effectively saying
$$\forall x \in \mathbb{Z}\  \forall y \in \mathbb{Z}\ \Big(\exists k \in \mathbb{Z}\ \big(x = 2k\big) \land \exists k \in \mathbb{Z}\ \big(y = 2k\big) \rightarrow \exists k \in \mathbb{Z}\ \big(x+y = 2k\big)\Big)$$
I know when I first started writing proofs, I would often get confused about why things were universally quantified, when the word 'all' wasn't explicitly mentioned. But over time, you learn to understand the implicit quantification, and in this statement, it is exactly meant to be true for any two even integers.

Also the proof of the statement is easy: Supposing $x$ and $y$ are even, then we can write them as $x = 2k_1$ and $y = 2k_2$ for some integers $k_1$ and $k_2$. So their sum is of course $x+y = 2k_1 + 2k_2$ $= 2(k_1 + k_2)$. But we know any integer plus itself is again an integer, so $x+y = 2k_3$ for an integer $k_3 = k_2 + k_1$. Of course since we assumed nothing about $x$ and $y$ other than the fact that they were even, we have proven the statement for all pairs of even integers.
A: Strictly speaking, your statement is not an implication, but it can be expressed as a quantified formula in which the universal quantifier governs a logical implication.
Let $P(x)$ mean "$x$ is even". Then your statement can be written as
$$
\forall n, k \in \mathbb{Z} \quad \left(\, P(n) \land P(k) \implies P(n+k) \,\right).
$$
The clause governed by the universal quantifier has the structure you want. The hypothesis is $P(n) \land P(k)$ and the conclusion is $P(n+k)$.
This is an example of a formula in a first-order logic. In simple terms, first-order logic extends propositional calculus which you are familiar with by adding quantifiers (such as $\forall$ meaning "for all"), predicates (such as $P(x)$ which we defined here to mean "$x$ is even"), functions (such as $+$) and constants (such as $0$ which we did not use here).
The statement can be proven for example in Peano's arithmetic.
A: It could be the case that you've only studied propositional (also called zeroth-order) logic so far. If so, then we may reword your statement slightly and write something like "if $p$ is even and $q$ is even, then $p+q$ is even too".
Nevertheless, such a formalution may feel lacking in some sense. I believe the culprit here is the word "always". To me this suggests something else is at play. The statement as you give it can't really be fully captured using propositional logic, because it contains an implicit quantifier. This means we have to use predicate logic (also called first-order logic) here. Note that what your statement really says is that for any even integers $p$ and $q$, $p+q$ is also even. Notationally, "for all $x$" is written as $\forall x$. Let's write $E(x)$ to express the property of being even. Hence, in formal notation, your statement becomes: $$\forall p \forall q \big(E(p)\land E(q)\to E(p+q)\big).$$
A: Here is a formal proof ... with the comments making up a mathematical proof:

