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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 $xRy$ ?

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  • $\begingroup$ Yes............ $\endgroup$ Commented May 27, 2023 at 19:23
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    $\begingroup$ Yes to which (two) of the three questions raised? ;) $\endgroup$
    – Piita
    Commented May 27, 2023 at 19:28
  • $\begingroup$ Does this answer your question? If I want to avoid quantifiers? $\endgroup$ Commented May 27, 2023 at 19:37

1 Answer 1

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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 $$\forall x{,}y{\in}A\;\forall z{\in}B\;\big(xRz\land yRz\implies x=y\big).\tag2$$
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