# Does $(x\in A\land y\in B)$ follow from $xRy$?

I have defined “being in $$R$$-relation to” as:

• Given a binary relation $$R\subseteq A\times B$$ and the elements $$x\in A$$ and $$y\in B,$$ $$xRy\iff (x,y)\in R.$$

For $$R$$ to be injective, is $$(xRz\land yRz)\implies x=y$$ acceptable or is $$\forall x, y, z\;\big((x,y\in A\land z\in B \land xRz\land yRz)\implies x=y\big)$$ necessary? But doesn't $$(x\in A\land y\in B)$$ follow from $$xRy$$ ?

• Yes............ May 27 at 19:23
• Yes to which (two) of the three questions raised? ;) May 27 at 19:28
• Does this answer your question? If I want to avoid quantifiers? May 27 at 19:37

For $$R$$ to be injective, $$\forall x, y, z\;\big((x,y\in A\land z\in B \land xRz\land yRz)\implies x=y\big)\tag1$$ But doesn't $$(x\in A\land y\in B)$$ follow from $$xRy$$ ?
Yes, since that membership condition is required for $$R$$'s definition to even be applicable. So, yes you can hide that condition, as well as the universal quantifiers, letting them be tacit, as you suggested:
$$(xRz\land yRz)\implies x=y\tag3$$
Notice that the statement $$(1)$$ is equivalent to this abbreviation, which you may prefer:
• given sets $$A$$ and $$B,$$ the relation $$R\subseteq A\times B$$ is injective if $$\forall x{,}y{\in}A\;\forall z{\in}B\;\big(xRz\land yRz\implies x=y\big).\tag2$$