Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to prove that the arithmetic axioms are independent by constructing a model in which all bar one of the axioms are satisfied, for each of the axioms below.

A first order theory with arithmetic has equality, one constant ($\mathbf{0}$), one unary function ('successor') and two binary functions ($+,\cdot$).

  1. $(\forall x)( x^+\neq0)$
  2. $(\forall x)(\forall y)(x^+=y^+\implies x=y)$
  3. $(\forall x)(x+0=x)$
  4. $(\forall x)(\forall y)(x+y^+=(x+y)^+)$
  5. $(\forall x)(x\cdot 0=0)$
  6. $(\forall x)(\forall y)(x\cdot (y^+)=(x\cdot y)+x)$
  7. Induction

I would appreciate some help constructing models in which the 3rd and 5th axioms are false in particular.

share|cite|improve this question
up vote 4 down vote accepted

Consider $\{0,1,\ldots\}$ with the same $+$, ${}^+$ and $0$, but a different multiplication: $a \cdot b = 1 + ab$ (where $ab$ is the usual product). This satisfies all but (5).

share|cite|improve this answer

All but 3: On $\mathbb N_0$, define $x+ y=(x+_{\text{standard}}y)^+$. Then use 5 and 6 to define multiplication recursively. Actually, one verifies that this results in $x\cdot y=x^+\cdot_{\text{standard}}y$

share|cite|improve this answer

In the same fashion of Robert's answer, redefine $a+b=a^+ + b$. This should violate the third axiom.

share|cite|improve this answer
Actually it also violates 4th and 6th axioms – mattecapu Apr 29 '14 at 9:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.