Are these statements about even numbers called symmetrical statements? I have these following statements.
x is a even number $\Rightarrow$ xy is a even  number
y is a even number $\Rightarrow$ xy is a even number
Can I call them symmetrical statements? 
 A: Usually, I use the word "symmetric" like this: I would say of the single statement
$$xy\text{ is even }\implies \text{ either }x\text{ is even, or }y\text{ is even}$$
that "it is symmetric in $x$ and $y$", because of the commutativity of multiplication. In this sense, even though it is true that when we switch the positions of $x$ and $y$ in the statement
$$x\text{ is even }\implies xy\text{ is even}$$
the resulting statement
$$y\text{ is even }\implies xy\text{ is even}$$
is true, I would not call the original statement "$x\text{ is even }\implies xy\text{ is even}$" symmetric in $x$ and $y$, because the meaning of the statement is changed when we switch the positions of $x$ and $y$.
Now, I don't think I usually hear a pair of statements, taken together, being referred to as "symmetric" or "symmetrical", but nevertheless I think it is clear enough that anyone would essentially know what you mean when you say it.
A: I haven't heard of these types of statements described as "symmetrical" before, but that it is an understandable way to put it.  One phrase I have heard in this context is "without loss of generality".  For example, suppose we have the lemma 
x is even ⟹ xy is even, and we also have two numbers x and y, at least one of which is even.  Then we might say "Without loss of generality, assume x is even.  Then therefore xy is also even."  The "WLoG" phrase draws attention to the symmetry without requiring explicit statements of both versions of the lemma.
A: I often read phrases like:

$\forall x\ y(\operatorname{even}(x)\to\operatorname{even}(x\cdot y))$ because …
$\forall x\ y(\operatorname{even}(y)\to\operatorname{even}(x\cdot y))$ holds by
  symmetry.

