Applying the definition of $<$, the theorem becomes: Exactly one of the following is true: :$(1): \quad a = b$ :$(2): \quad \exists n \in \N_{>0} : b + n = a$ :$(3): \quad \exists n \in \N_{>0} : a + n = b$ We will use the principle of Mathematical Induction. Let $P \left({a}\right)$ be the proposition that exactly one of the above is true, for all natural numbers $b$, for fixed natural number $a$. === Basis for the Induction === We will prove that the proposition is true for $a = 0$. Using Proof by Cases, we divide the proposition into two cases. ==== Case 1: $b = 0$ ==== ===== $(1)$ is true ===== It follows trivially from the values of $a$ and $b$. ===== $(2)$ is false ===== {{AimForCont}} $c$ is a non-zero natural number such that: :$b + c = a$ Substituting the values of $a$ and $b$, we obtain: :$0 + c = 0$ By Natural Number Addition Commutes with Zero, we can simplify the equation to: :$c = 0$ which is a contradiction of the assumption that $c$ is non-zero. ===== $(3)$ is false ===== {{AimForCont}} $c$ is an non-zero natural number such that: :$a + c = b$ Substituting the values of $a$ and $b$, we obtain: :$0 + c = 0$ By Natural Number Addition Commutes with Zero, we can simplify the equation to: :$c = 0$ which is a contradiction of the assumption that $c$ is non-zero. ==== Case 2: $b \ne 0$. ==== ===== $(1)$ is false ===== It follows trivially from the fact that $a = 0$. ===== $(2)$ is false ===== {{AimForCont}} $c$ is an non-zero natural number such that: :$b + c = a$ Substituting the values of $a$, we obtain: :$b + c = 0$ By Non-Successor Element of Peano Structure is Unique, there exists a natural number $d$ such that: :$s \left({d}\right) = c$ where $s$ denotes the successor mapping. Therefore, we have: :$b + s \left({d}\right) = 0$ Applying the definition of addition in Peano structure, we get: :$s \left({b + d}\right) = 0$ which is a contradiction of $(P4)$: $0$ is not in the image of $s$. ===== $(3)$ is true ===== By Natural Number Addition Commutes with Zero, we have: :$a + b = b$ The result follows. === Inductive hypothesis === It is now assumed that the proposition is true for $a = k$, where $k$ is a natural number. That is, for all natural numbers $b$, exactly one of the following is true: :$(1): \quad k = b$ :$(2): \quad k > b$ :$(3): \quad k < b$ Then, it is to be proved that the proposition is true for $a = s \left({k}\right)$. That is, for all natural numbers $b$, exactly one of the following is true: :$(1'): \quad s \left({k}\right) = b$ :$(2'): \quad s \left({k}\right) > b$ :$(3'): \quad s \left({k}\right) < b$ === Inductive step === We have: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = s \left({k + 0}\right)       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | r = k + s \left({0}\right)       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | r = k + 1       | c = {{Defof|One}} }} {{end-eqn}} ==== Case 1: $k = b$ ==== In this case: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = k + 1 }} {{eqn | r = b + 1 }} {{end-eqn}} ===== $(1')$ is false ===== {{AimForCont}} $(1')$ is true. Then: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = b       | c = Assumption }} {{eqn | l = b + 1       | r = b       | c =  }} {{eqn | l = 1       | r = 0       | c = Natural Number Addition is Cancellable }} {{eqn | l = s \left({0}\right)       | r = 0       | c = {{Defof|One}} }} {{end-eqn}} which is a contradiction of $(P4)$: $0$ is not in the image of $s$. ===== $(2')$ is true ===== This is apparent from the definition of $>$. ===== $(3')$ is false ===== {{AimForCont}} $(3')$ is true. Let $c$ be a non-zero natural number such that: :$s \left({k}\right) + c = b$ Then: {{begin-eqn}} {{eqn | l = s \left({k}\right) + c       | r = b       | c = Assumption }} {{eqn | l = \left({b + 1}\right) + c       | r = b       | c =  }} {{eqn | l = b + \left({c + 1}\right)       | r = b       | c = Natural Number Addition is Associative and Natural Number Addition is Commutative }} {{eqn | l = c + 1       | r = 0       | c = Natural Number Addition is Cancellable }} {{eqn | l = c + s \left({0}\right)       | r = 0       | c = {{Defof|One}} }} {{eqn | l = s \left({c}\right)       | r = 0       | c = {{Defof|Addition in Peano Structure}} }} {{end-eqn}} which is a contradiction of $(P4)$: $0$ is not in the image of $s$. ==== Case 2: $k < b$ and $s \left({k}\right) = b$ ==== ===== $(1')$ is true ===== This follows from the assumption. ===== $(2')$ is false ===== {{AimForCont}} $(2')$ is true. Let $c$ be a non-zero natural number such that: :$s \left({k}\right) = b + c$ Then: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = b + c       | c = Assumption }} {{eqn | l = b       | r = b + c       | c = Assumption }} {{eqn | l = 0       | r = c       | c = Natural Number Addition is Cancellable }} {{end-eqn}} which is a contradiction of our assumption that $c$ is non-zero. ===== $(3')$ is false ===== {{AimForCont}} $(3')$ is true. Let $c$ be a non-zero natural number such that: :$s \left({k}\right) + c = b$ Then: {{begin-eqn}} {{eqn | l = s \left({k}\right) + c       | r = b       | c = Assumption }} {{eqn | l = b + c       | r = b       | c = Assumption }} {{eqn | l = c       | r = 0       | c = Natural Number Addition is Cancellable }} {{end-eqn}} which is a contradiction of our assumption that $c$ is non-zero. ==== Case 3: $k < b$ and $s \left({k}\right) \ne b$ ==== From the definition of $<$, we have the existence of a non-zero natural number $m$ such that: :$k + m = b$ By Non-Successor Element of Peano Structure is Unique, there exists a natural number $p$ such that: :$s \left({p}\right) = m$ Substituting: :$k + s \left({p}\right) = b$ Applying the definition of addition in Peano structure, we get: :$s \left({k + p}\right) = b$ ===== $(1')$ is false ===== It is assumed in this case that: :$s \left({k}\right) \ne b$ Therefore, $(1')$ is false. ===== $(2')$ is false ===== {{AimForCont}} $(2')$ is true. Let $c$ be a non-zero natural number such that: :$s \left({k}\right) = b + c$ By Non-Successor Element of Peano Structure is Unique, there exists a natural number $d$ such that: :$s \left({d}\right) = c$ Then: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = b + c       | c = Assumption }} {{eqn | r = \left({k + s \left({p}\right)}\right) + s \left({d}\right)       | c =  }} {{eqn | r = k + \left({s\left({p}\right) + s \left({d}\right)}\right)       | c = Natural Number Addition is Commutative }} {{eqn | r = k + s\left({p + s \left({d}\right)}\right)       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | r = s\left({k + \left({p + s \left({d}\right)}\right)}\right)       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | r = s\left({k + \left({s \left({p + d}\right)}\right)}\right)       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | l = k       | r = k + \left({s \left({p + d}\right)}\right)       | c = $(P3)$$:$ $s$ is injective }} {{eqn | l = 0       | r = s \left({p + d}\right)       | c = Natural Number Addition is Cancellable }} {{end-eqn}} which is a contradiction of $(P4)$: $0$ is not in the image of $s$. ===== $(3')$ is true ===== We have: {{begin-eqn}} {{eqn | l = s \left({k + p}\right)       | r = b       | c =  }} {{eqn | l = s \left({p + k}\right)       | r = b       | c = Natural Number Addition is Commutative }} {{eqn | l = p + s \left({k}\right)       | r = b       | c = {{Defof|Addition in Peano Structure}} }} {{eqn | l = s \left({k}\right) + p       | r = b       | c = Natural Number Addition is Commutative }} {{end-eqn}} The result follows. ==== Case 4: $k > b$ ==== From the definition of $>$, we have the existence of a non-zero natural number $m$ such that: :$k = b + m$ By Non-Successor Element of Peano Structure is Unique, there exists a natural number $p$ such that: :$s \left({p}\right) = m$ Substituting: :$k = b + s \left({p}\right)$ By applying $s$ to both sides, we obtain: :$s \left({k}\right) = s \left({b + s \left({p}\right)}\right)$ Applying the definition of addition in Peano structure, we get: :$s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$ ===== $(1')$ is false ===== {{AimForCont}} $(1')$ is true. Then: {{begin-eqn}} {{eqn | l = s \left({k}\right)       | r = b       | c = Assumption }} {{eqn | l = b + s \left({s \left({p}\right)}\right)       | r = b       | c =  }} {{eqn | l = s \left({s \left({p}\right)}\right)       | r = 0       | c = Natural Number Addition is Cancellable }} {{end-eqn}} which is a contradiction of $(P4)$: $0$ is not in the image of $s$. ===== $(2')$ is true ===== We have: :$s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$ The result follows. ===== $(3')$ is false ===== {{AimForCont}} $(3')$ is true. Let $c$ be a non-zero natural number such that: :$s \left({k}\right) + c = b$ Then: {{begin-eqn}} {{eqn | l = s \left({k}\right) + c       | r = b       | c = Assumption }} {{eqn | l = \left({b + s \left({s \left({p}\right)}\right)}\right) + c       | r = b       | c =  }} {{eqn | l = b + \left({c + s \left({s \left({p}\right)}\right)}\right)       | r = b       | c = Natural Number Addition is Associative and Natural Number Addition is Commutative }} {{eqn | l = \left({c + s \left({s \left({p}\right)}\right)}\right)       | r = 0       | c = Natural Number Addition is Cancellable }} {{eqn | l = s \left({c + s \left({p}\right)}\right)       | r = 0       | c = {{Defof|Addition in Peano Structure}} }} {{end-eqn}} which is a contradiction of $(P4)$: $0$ is not in the image of $s$. {{qed}}