By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers. To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$. We have: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {c, d} {}       | o = \not <       | r = \eqclass {a, b} {}       | c = Strict Ordering on Integers is Asymmetric }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \prec       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d       | o = \ne       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = \ne       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = >       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {c, d} {}       | o = \not >       | r = \eqclass {a, b} {}       | c = Strict Ordering on Integers is Asymmetric }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = >       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \succ       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d       | o = \ne       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = \ne       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = =       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = =       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d       | o = \not \prec       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | lo= \land       | l = a + d       | o = \not \succ       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = \not <       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | lo= \land       | l = \eqclass {a, b} {}       | o = \not >       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} This demonstrates that $<$, $=$ and $>$ are mutually exclusive. Now: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \prec       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = b + c       | o = \preccurlyeq       | r = a + d       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {c, d} {}       | o = \le       | r = \eqclass {a, b} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = >       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \succ       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = b + c       | o = \succcurlyeq       | r = a + d       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {c, d} {}       | o = \ge       | r = \eqclass {a, b} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = \ne       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \ne       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d       | o = \prec       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | lo= \lor       | l = a + d       | o = \succ       | r = b + c       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | lo= \land       | l = \eqclass {a, b} {}       | o = \>       | r = \eqclass {c, d} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} demonstrating that either $<$, $=$ or $>$ must hold. {{qed}}