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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.

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3 Answers 3

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).

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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 $x\cdot y=x^+\cdot_{\text{standard}}y$

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

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Actually it also violates 4th and 6th axioms –  mattecapu Apr 29 at 9:57

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