I If symmetric then transitive relation

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The discussion explores the relationship between symmetric and transitive properties in relations. It argues that while symmetry (if xRy and yRx, then xRx) suggests a form of transitivity, it is not sufficient to establish true transitivity, which requires a broader set of elements. An example is provided where a relation is symmetric and reflexive but not transitive, demonstrating that transitivity can necessitate at least three distinct elements. The conversation also highlights that even in transitive relations, such as the order relation '>', the conditions for symmetry may not hold. Overall, the nuances of these relational properties are examined, emphasizing their distinct definitions and requirements.
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Isn't, if we have xRy and yRx then xRx will also make transitive? Because if I am right {(x,x),(y,y)} on set {x,y} is symmetric and transitive.

Isn't the above similar to, if xRy and yRz then xRz is transitive relation?

Thanks.
 
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##R = \{(x,x),(y,y),(z,z),(x,y),(y,x),(y,z),(z,y)\}##
is an example of a relation that's symmetric and reflexive without being transitive, because
##xRy \leftrightarrow yRx##
and
##yRz \leftrightarrow zRy##
but we also have ##xRy \wedge yRz## without ##xRz##.
 
Similar but not nearly as strong. xRy and yRx => xRx is a statement about a much smaller set of x and y than the transitive property requires.
 
FactChecker said:
Similar but not nearly as strong. xRy and yRx => xRx is a statement about a much smaller set of x and y than the transitive property requires.
Does this mean transitive relation require atleast 3 distinct element of a set e.g {x,y,z}.

Also as I mentioned, is {(x,x),(y,y)} on set {x,y} reflexive along with symmetric and transitive.
 
(xRy and yRx) => xRx only makes a statement about the x & y where both xRy and yRx. There might easily be none of those, so it might say nothing.

PS. Even if the relation R is transitive, there may be no x & y where (xRy and yRx). An example is the order relation '>'. It's not possible for (x > y & y > x), even though '>' is a transitive relation.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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