# First order logic: Showing that a bijection preserves homomorphism

The symbol $$0^{mathbb R}$$ is a 0-ary function symbol, and you need to handle it as well. Since $$-0 = 0$$, this is not a problem.

As for $$+^{mathbb R}$$, high school arithmetic shows that
$$-(x+y) = (-x) + (-y).$$

More generally, $$-$$ is an automorphism of any Abelian group. An Abelian group is a set together with a binary operator $$+$$, a unary operator $$-$$, and a constant $$0$$ such that:

1. $$x+y = y+x$$.
2. $$(x+y)+z = x+(y+z)$$.
3. $$x+0 = x$$.
4. $$x+(-x) = 0$$.

The mapping $$x mapsto -x$$ is an automorphism. Notice that

• For all $$x$$, $$(-x)+x=x+(-x)=0$$.
• For all $$x$$, $$(-x)+(-(-x))=0$$.
• Hence $$x = x+0 = x+((-x)+(-(-x))) = (x+(-x))+(-(-x)) = 0+(-(-x)) = (-(-x))+0 = -(-x)$$.

Therefore $$-$$ is an involution. In particular, if $$-x=-y$$ then $$x=-(-x)=-(-y)=y$$, so that $$-$$ is one-to-one. Since $$-(-x)=x$$, it is also onto.

Since $$0 = 0 + (-0) = (-0) + 0 = -0$$, we see that $$-$$ maps $$0$$ to itself.

Since $$-(-x) = -(-x)$$, we see that $$-$$ maps $$-$$ to itself.

Finally,
$$(x+y)+((-x)+(-y)) = (y+x)+((-x)+(-y)) = ((y+x)+(-x))+(-y) = (y+(x+(-x)))+(-y) = (y+0)+(-y) = y+(-y) = 0.$$
Therefore
$$-(x+y) = (-(x+y)) + 0 = (-(x+y)) + ((x+y)+((-x)+(-y))) = ((-(x+y))+(x+y)) + ((-x)+(-y)) = ((x+y)+(-(x+y))) + ((-x)+(-y)) = 0 + ((-x)+(-y)) = ((-x)+(-y)) + 0 = (-x)+(-y).$$
This shows that $$-$$ maps $$+$$ to itself.