# 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?

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I can't make anything of this question... – Evariste Oct 1 '11 at 20:10
I'm sorry my fault. "Equal" and "even" has the same name in my native language. – user145 Oct 1 '11 at 20:21
Please use more specific titles. General areas like "logic" are handled using tags. Whether the question is easy is subjective and doesn't tell the reader anything about what the question is about. A more specific title might be for instance: Are these statements about even numbers called symmetrical statements? – joriki Oct 1 '11 at 20:27
Ok. I have change the title now. – user145 Oct 1 '11 at 20:30

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.

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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.

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$\forall x\ y(\operatorname{even}(x)\to\operatorname{even}(x\cdot y))$ because …

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

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