I am going to try something along Peano's thinking

[Johnny5]

If I make an error in what follows, I would like it pointed out.

 

The purpose has something to do with 0!, and 0^0, but I am not going to say what.

 

Preliminary work: There is a superset [math] \mathbb{S} [/math], and any element of that superset is called a number.

 

Axiom A

 

[math] 0 \in \mathbb{S} [/math]

 

Axiom B

 

[math] \forall x \in \mathbb{S}[0+x=x] [/math]

 

Axiom C

 

[math] \forall x \in \mathbb{S}\forall y \in \mathbb{S} [x+y \in \mathbb{S}] [/math]

 

Axiom D

 

[math] \mathbb{N} \subset \mathbb{S} [/math]

 

Undefined binary relation on S: <

 

Definition:

 

[math]\forall x,y \in \mathbb{S} [ x > y \Leftrightarrow y<x ] [/math]

 

Definition (equality of two numbers):

 

[math]\forall x,y \in \mathbb{S} [ x = y \Leftrightarrow not(x<y) \ \& \ not(y<x) ] [/math]

 

 

 

Axiomatic development of the set of natural numbers

 

Axiom I

 

[math] 1 \in \mathbb{N} [/math]

 

Axiom II

 

[math] \forall x[x \in \mathbb{N} \Rightarrow x+1 \in \mathbb{N}] [/math]

 

Theorem: [math] 1+1 \in \mathbb{N} [/math]

 

Proof:

 

By axiom 1, 1 is an element of the set of natural numbers. Now, plug 1 into axiom two to get the following true statement:

 

[math] 1 \in \mathbb{N} \Rightarrow 1+1 \in \mathbb{N} [/math]

 

The antecedent of the statement above, is axiom 1, hence true. Therefore, the consequent is also unconditionally true(modus ponens). Therefore:

 

[math] 1+1 \in \mathbb{N} [/math]

 

Which is the theorem. QED

 

It isn't hard to see that using the theorem just proven, together with axiom2, that 1+1+1 is a natural number. And we can keep repeating the same steps ad infinitum. So that much is clear at this point.

 

However, there has not been introduced enough axioms to reach the following conclusion yet:

 

not (1=1+1)

 

And similarly we will want to know that not (1+1=1+1+1), and so on.

 

Axiom III:

[math] \forall y \in \mathbb{N} [ y < y+1] [/math]

 

The axiom above is going to be true, regardless of whether we take it as axiom, or theorem.

 

From it, it now follows that 1<1+1, 1+1<1+1+1, etc.

And also, it now follows that not(1=1+1), not(1+1=1+1+1), etc.

 

Since 1<1+1, it follows from the definition of equality of two numbers, that it cannot be the case that 1=1+1, because if 1=1+1 then not (1<1+1), etc.