I have some confusion when it comes time to instantiate quanitfiers, the specific case is this:
$$\forall x\forall y((\exists z(Rxz\land Rzy))\Rightarrow Rxy)\space\space\space(1)$$
My understanding is that because existential quantifiers instantiate an arbitrary variable, I can just substitute any filler variable for both x and y, which leads to,
$$\exists z(Roz\land Rzo)\Rightarrow Roo\space\space\space(2)$$
But this seems to obfuscate the x and y variables, and I could work backwards and generalize:
$$\forall x(\exists z(Rxz\land Rzx)\Rightarrow Rxx)\space\space\space(3)$$
Which seems obviously wrong to me. Ultimately my question is whether both (2) and (3) are correct, (2) is but (3) is not, or neither are, and what exactly is the flaw in my understanding?
(2) and (3) are both consequences of (1). If the statement is true for all $x$ and $y$, it is true in the cases when $y=x$ and $x$ is anything.
Note that I said (2) and (3) are consequences of (1) and not equivalent to it . (1) is the statement that $R$ is transitive. Statement $3$ is a weaker property. We wouldn't be able to use this weaker property to conclude $1R3$ from $1R2$ and $2R3$ but we could use it to conclude $1R1$ from $1R2$ and $2R1$. However we could conclude either from full transitivity (1).