I need to find a flaw in the proof of the following statement:
Any relation that is symmetric and transitive is also reflexive. False Proof: aRb implies bRa,by symmetry. Then by transitivity, aRa holds. Where is the flaw? And also, what is the type of fallacy?
The flaw is basically that: $~P\to Q ~\nvdash~ P\wedge Q$
Thus in the case: $$\forall a{\in}S~\forall b{\in}S~(\color{blue}{aRb\to bRa}), \forall a{\in}S~\forall b{\in}S~(\color{navy}{aRb \wedge bRa} \to aRa) ~\nvdash~ \forall a{\in}S~(aRa)$$
Take for instance the set $S=\{x,y,z\}$ and a relation $R=\{(y,y),(y,z),(z,y),(z,z)\}$. The relation $R$ is symmetric and transitive, but it is not reflexive (on $S$).