# Is there any way to prove that $x = y \Rightarrow x + z = y + z$?

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?

• 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. – 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 $$y+z = x+z$$