# Using Peano's axioms to disprove the existence of self-looping tendencies in natural numbers

Let me clarify by what I mean by "self-looping". So, we know that Peano's axioms use primitive terms like zero, natural number and the successor operation. Now, I want to prove that the following isn't a valid interpretation of the axioms:

$$succ(0) = 1\qquad succ(1) = 2\qquad succ(2) = 3\qquad succ(3) = 3$$

This is a system that is "self-looping" at the number three and I want to disprove its existence using the axioms. So, I'll use the axiom that $$succ(n) = succ(m)\implies n=m$$. Applying this rule, I get that $$succ(3) = succ(2)\implies 3=2$$. Now, to show that the axiom isn't obeyed, I must prove that $$3\neq2$$. And, from here, I'm not sure what to do. I can try using transitivity to say the following by first assuming that $$2=3$$: $$2=3\implies succ(1) = 3$$ From the above four equations that define the $$succ$$ operator it is evident that, $$succ(1)\neq3$$. Therefore, $$2\neq3$$. Just wondering if this proof is right. I ask because the textbook that I am referring to, Tao's Analysis I, uses a slightly longer proof, which, to me, seems unnecessary (this isn't the actual proof he writes but the method is analogous to what he wrote in the book). Tao first proves that $$1\neq0$$ using the axiom that for all natural numbers, $$succ(n)\neq0$$: $$1=succ(0)\neq0$$ Now, he writes $$2=3$$ as $$succ(1)=succ(2)$$. By the injectivity axiom, this implies that $$1=2$$. He further writes $$succ(0) = succ(1)$$. Again, by the injectivity axiom, $$0=1$$. This, we saw, wasn't true. Therefore, $$2\neq3$$.

So, I'm wondering whether my proof is valid. Or does Tao's proof address something that I'm missing? Thanks.

• @MauroALLEGRANZA could you please explain why the proof I provided isn't sufficient? Commented May 7 at 8:44
• @MauroALLEGRANZA Also, Tao did not use induction to make the proof. He discusses this methodology before introducing the mathematical principle of induction. So isn't that considered Robinson arithmetic? Commented May 7 at 8:46

## 1 Answer

From the above four equations that define the $$succ$$ operator it is evident that $$succ(1)\neq3$$

Why? Because $$succ(1) = 2$$? But don't you need $$2 \neq 3$$ in the first place to claim that this leads to $$succ(1) \neq 3$$? Your logic would be circular.

• I see! That's a subtle point. Thanks for the clarification Commented May 9 at 10:53