NT Insights Vol 1.1 Addition on Natural Numbers

We have already seen how the 5 axioms together generate the entire set of natural numbers $\mathbb{N}$ in Vol 1.0. Now we will define a binary operation, called addition, on $\mathbb{N}$ in terms of those 5 axioms.
Before that, let us give these elements of $\mathbb{N}$ their usual names, as follows:

1. $s(0) = 1$
2. $s(s(0)) = s(1) = 2$
3. $s(s(s(0))) = s(s(1)) = s(2) = 3$
   and so on…

An important observation is that the successive natural numbers differ by a single reference to $s()$. Let us call a reference to $s()$ as a unit distance. Then, clearly, the successive natural numbers are unit distance apart. Also, the difference between any two natural numbers is a multiple of this unit distance.
Another important observation is that $0$ does not involve any reference to function $s()$. This is so because all other numbers are successors of $0$, but $0$ is not successor of any natural number (Reminds me of Lord Shiva, who is called Anaadi). The magnitude of distance, therefore, can not be assigned to $0$. Whereas, for every non-zero natural number $b$, its distance from $0$ = number of calls to $s()$ needed to produce $b$.

Thus, when we say that a natural number is $b$, we actually mean that this natural number is $b$ units away from $0$.

We observe that every successor of $0$ derives its magnitude (or value), relative to $0$, the magnitude of $0$ itself is null ($\because 0$ is the reference point or origin of all its successors). A consequence of this is that, when we include the magnitude (distance) of $0$ into distance of another natural number, say $b$, then the total distance still is equal to $b$. This leads us to define addition as follows:

Addition: Define a mapping $+ : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ as

$a + 0 := a\ \forall a\in \mathbb{N} \tag{1.0}$
$a + s(b) := s(a+b)\ \forall a,b\in \mathbb{N} \tag{1.1}$

Putting $b=0$ in (1.1), we get $a+s(0) = a+1 = s(a+0)= s(a)$.
Thus, $s(a) = a+1\ \forall a\in \mathbb{N}$, which is as expected because successive natural numbers are unit distance apart.
Notice that:
$2:=s(1)$ is a definition, whereas
$1+1 = 1+s(0)=s(1+0)=s(1)$ is a theorem (derived by applying definitions).

Proposition 1.2: Equations (1.1) and (1.2) define addition $(m+n)$ for all $m,n\in \mathbb{N}$.
Proof: We need to show that the mapping $(m,n) \to (m+n)$ is defined $\forall\ (m,n) \in \mathbb{N}\times \mathbb{N}$. Consider an arbitrary natural number $m\in \mathbb{N}$. If we show that $(m+n)$ is definied for all $n\in \mathbb{N}$, then we are done ($\because$ m being arbitrary, spans entire $\mathbb{N}$).
But, we have to prove this by using the 5 axioms and nothing else. The property $(m+n)\in \mathbb{N}$ has to be satisfied for all $n\in \mathbb{N}$. So let us consider the set of all $n$ satisfying this property. Define $S=\{ n\in \mathbb{N}: (m+n)\in \mathbb{N} \}$. Since we have to prove the result for all $n\in \mathbb{N}$, we can just show that $S = \mathbb{N}.$
To show this equivalence, inductive axiom ($5^{th}$ axiom) can be employed.
$0\in S$ $(\because m+0 \in \mathbb{N}$ by definition (1.0)).
Assume $n\in S \implies (m+n)\in \mathbb{N}$.
Then, $m+(n+1) = m + s(n) = s(m+n) \in \mathbb{N}$ ($\because (m+n)\in \mathbb{N}$; use invariance of $s()$).
$\implies (n+1) \in S$
Thus, by inductive axiom of Peano, $S = \mathbb{N}$. $\space\blacksquare$

Proposition 1.3 (Associativity of Addition): For all $a,b,c \in \mathbb{N}$, we have $a+(b+c) = (a+b)+c$.
Proof: Assume $a, b \in \mathbb{N}$ to be arbitrary, i.e., (a,b) spans $\mathbb{N} \times \mathbb{N}$. Define set
$S=\{ c\in \mathbb{N}: a+(b+c) = (a+b)+c \}$.
Now if we show that $S= \mathbb{N}$, then we are done (Compare this proof strategy with that in Proposition 1.2).
$0 \in S$, because
LHS: $(a+b)+0 = (a+b)$ and RHS: $a+(b+0) = a+(b)$
$\implies$ LHS=RHS..
Assume $c\in S \implies a+(b+c) = (a+b)+c$.
Then, LHS: $a+(b+s(c))=a+s(b+c)=s(a+(b+c))$ and
RHS: $(a+b)+s(c)=s((a+b)+c)=s(a+(b+c))\ \because (c\in S)$.
$\implies$ LHS=RHS.
Hence, by induction axiom of Peano, $S=\mathbb{N}$. $\space\blacksquare$

By definition we know that $a+0 = a$; since we haven’t proved commutativity of addition yet, we don’t know that $0 + a = a$ at this point.
Lemma 1.4: For all $a\in \mathbb{N}$, $0+a=a$.
Proof: Following previous strategies, we define
$S=\{ a\in \mathbb{N}: 0+a=a \}$. Now we need to show that $S=\mathbb{N}$.
Observe that, $0+0=0$. So, $0\in S$.
Assume that $a\in S \implies 0+a=a$.
Then, $0+s(a)=s(0+a)=s(a)$ ($\because a\in S$).
$\implies s(a)\in S$.
Hence, by inductive axiom, $S=\mathbb{N}$. $\space\blacksquare$

Lemma 1.5 For all $x,y\in\mathbb{N}$, we have $s(x)+y=x+s(y)$.
Proof: Consider arbitrary $x\in\mathbb{N}$. Define the set
$S=\{ y\in\mathbb{N}:s(x)+y=x+s(y) \}$.
Now we need to show that $S=\mathbb{N}$.
$x+s(0)=s(x+0)=s(x)=s(x)+0$.
Thus, $0\in S$.
Assume, $y\in S$, i.e., $s(x)+y=x+s(y)$.
Then, $s(x)+s(y)=s(s(x)+y)=s(x+s(y))$ ($\because y\in S$).
$=x+s(s(y))$.
$\implies s(x)\in S$.
Thus, by inductive axiom, $S=\mathbb{N}$. $\space\blacksquare$