Why does $\exists$ distribute over conjunction here?

Question

Here I used:

$$\small \exists y:\exists z:x=(y,z)\wedge (y,z)\in R \wedge (y,z)\in R\leftrightarrow\exists y:\exists z:x=(y,z)\wedge\exists z:(y,z)\in R\wedge\exists z:(y,z)\in R$$ But why does $\exists$ distribute over conjunction here?

Answer 1

Typically, $\exists$ does not distribute over conjunctions, but in your specific case it (mostly) does because:

$\exists y : \exists z : (x = (y,z) \land (y,z) \in R \land (y,z) \in R )\Leftrightarrow$

$\exists y : \exists z :( x = (y,z) \land x \in R \land x \in R )\Leftrightarrow$ (Null Quantification!)

$\exists y : \exists z : (x = (y,z) \land \exists z : x \in R \land \exists z : x \in R) \Leftrightarrow$

$\exists y : \exists z : (x = (y,z) \land \exists z : (y,z) \in R \land \exists z : (y,z) \in R)$

though please note that this last statement is not quite the same statement as:

$\exists y : (\exists z : x = (y,z) \land \exists z : (y,z) \in R \land \exists z : (y,z) \in R)$

which is what the result would be of distributing the existential over the conjunction. In other words, it looks like you are distributing the existential over the conjunction, but it's actually a little different from that.