Implication and symmetric relations

Question

I read that the symmetric relation can be written with math logic symbols as an implication( (a,b)∈ R->(b,a)∈ R) ).So i thought that it should have the same truth table with the implication.But say S={1,2,3,4} and R={(1,1),(1,2),(2,1)},and take the case where (a,b)∉R->(b,a)∈R (FALSE->TRUE).If they have the same tables shouldn't to have the same value? (In this case TRUE) but if we search the SxS we see that the implication is FALSE where it sould be TRUE What i do not understand well? I hope i was clear thank you!

Answer 1

I read that the symmetric relation can be written with math logic symbols as an implication( (a,b)∈ R->(b,a)∈ R) ).

As an universal implication, to be precise.   It is $~\forall a{\in}S~\forall b{\in}S~\big((a,b)\in R~\to~(b,a)\in R\big)$, which says, "if any pair is $R$-related, then the inverse pair is $R$-related too".

take the case where (a,b)∉R->(b,a)∈R (FALSE->TRUE).

No, if you have that occur, then the relation is not symmetric.   Repeat: if any pair is in a symmetric relation, then its inverse is in the relation too.

So, if there is some $(b,a)$ in a symmetric relation, then its inverse, $(a,b)$, must be in the relation too.

Rather, the symmetry criteria is equivalent to $\forall a{\in}S~\forall b{\in}S~\big((a,b)\notin R \to (b,a)\notin R\big)$.

But say S={1,2,3,4} and R={(1,1),(1,2),(2,1)},

There is no pair that is in the relation without its inverse being there too.   Note: $(1,1)$ is its own inverse.

Likewise, there is no pair that is not in the relation, without its inverse being not in it either.