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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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    Ok. I have change the title now.2011-10-01

3 Answers 3

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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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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.