The case when
x<y will be proven. The other cases can be proven using a similar reasoning. Consider the before beginning the proof.
x<y⇔y−x>0
This property can be proven using the of
z, which is
-z. Now, the can be applied to the second part of the statement.
y−x>0
y−x+0>0
y−x+(-z−(-z))>0
y+(-z−(-z))−x>0
y−z−(-z)−x>0
y−z−(-z+x)>0
y−z−(x−z)>0
(y−z)−(x−z)>0
From the biconditional statement, the last inequality can be rewritten.
(y−z)−(x−z)>0⇕x−z<y−z
Finally, because
x<y, the property has been proven.
