Why is a symmetric relation defined: $\forall x\forall y( xRy\implies yRx)$ and not $\forall x\forall y (xRy\iff yRx)$?

If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $A\times A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(\forall x\in A)(\forall y\in A):x\mathrel Ry\implies y\mathrel Rx.\tag1$$And this is equivalent to$$(\forall x\in A)(\forall y\in A):x\mathrel Ry\iff y\mathrel Rx.\tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $A\implies B$ than $A\iff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.

For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).

1) $(x,y) \in R \implies (y,x) \in R$ for ALL $x,y \in A$

And the statement 2) $(x,y) \in R \iff (y,x) \in R$ are equivalent statements.

If 1) is true and $(x,y) \not \in R$ then although $(x,y)\in R\implies (y,x)\in R$ or $F \implies (y,x)\in R$ is true, it does not tell us any thing about whether or not $(y,x) \in R$. However $(y,x) \in R \implies (x,y) \in Y$ tells us that $(y,x) \not \in R$. Because $(y,x) \in R \implies (x,y) \in R$ means $(y,x) \in R \implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) \not \in R$.

So in your example you have $(1,2)\in R\implies (2,1)\in R$ is true but you don't have $(2,1) \in R \implies (1,2) \in R$ as true.

So it isn't symmetric.

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Another way to look at it:

If $A = \{1,2,3\}$

Then we will have 9 statments.

By 1) the nine statements are:

$(1,1)\in R\implies (1,1) \in R$

$(1,2) \in R \implies (2,1) \in R$

$(1,3) \in R \implies (3,1) \in R$

$(2,1) \in R \implies (1,2) \in R$

... etc... all nine are needed.

With 2) we also have nine statements:

$(1,1)\in R\iff (1,1) \in R$

$(1,2) \in R \iff (2,1) \in R$

$(1,3) \in R \iff (3,1) \in R$

$(2,1) \in R \iff (1,2) \in R$

...etc....

$(1,2) \in R \iff (2,1) \in R$ and $(2,1) \in R \iff (1,2)\in R$ is redundant.

So aesthetically, using definition 2) is .... inefficient.

If $xRy \implies yRx$ for all $x$ and all $y$, then we can choose $x := \tilde{y}$ and $y := \tilde{x}$ and get $\tilde{y}R\tilde{x} \implies \tilde{x}R\tilde{y}$ or, equivalently, $yRx \implies xRy$, which is $\impliedby$.

In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.