Terence Tao, Analysis I, 3e,

A.7 Equality

(...) How equality is defined depends on the class T of objects under consideration, and to some extent is just a matter of definition. However, for the purposes of logic we require that equality obeys the following four axioms of equality:

  • (Reflexive axiom) (...)
  • (Symmetry axiom) (...)
  • (Transitive axiom) (...)
  • (Substitution axiom). Given any two objects $x$ and $y$ of the same type, if $x = y$, then $f(x) = f(y)$ for all functions or operations $f$.

Concerning the substitution axiom, I keep wondering if there is really no way one could prove that

$$ x = y \Rightarrow x + z = y + z, $$

where $x, y, z$ are natural numbers?

  • 1
    $\begingroup$ Of course one can prove it, using other axioms of, say, real numbers. But this is not the way here. By the substitution axiom, we obtain this consequence with $f(x)=x+z$ without further ado. $\endgroup$ – Dietrich Burde Feb 14 at 21:19

$y+z = (y+0) + z = y+(x-x) + z = (y-x) + x+ z$

if $y = x$ then $y-x = 0$ and $y+z = x+z$

  • $\begingroup$ Is subtraction defined on natural numbers? $\endgroup$ – Max Herrmann Feb 14 at 21:33
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    $\begingroup$ @MaxHerrmann I was thinking about the field axioms for addition. The natural numbers are not a field. I'll have to think for a little bit more on how to do this without the field axioms. $\endgroup$ – Doug M Feb 14 at 22:30

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