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It is straightforward to show that the relation of equinumerousness is an equivalence relation: equinumerousness of ''A'' with ''A'' is witnessed by <math>i_A</math>; if ''f'' witnesses <math>|A|=|B|</math>, then <math>f^{-1}</math> witnesses <math>|B|=|A|</math>; if ''f'' witnesses <math>|A|=|B|</math> and ''g'' witnesses <math>|B|=|C|</math>, then <math>g\circ f</math> witnesses <math>|A|=|C|</math>.
It can be shown that <math>|A| \leq |B|</math> is a linear order on abstract cardinals, but not on sets. Reflexivity is obvious and transitivity is proven just as for equinumerousness. The [[
*<math>|A| \leq |B| \wedge |B| \leq |A| \rightarrow |A| = |B|</math>
(this establishes antisymmetry on cardinals), and
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