Produce a nonstandard model of Robinson arithmetic

Question

I wanted to know if my proof is correct, please, thanks for reading.

Robinson arithmetic, $\mathbf{Q}$, is a first-order theory with the following axioms:

$1.\space \neg\left(\exists x \left[S\left(x\right)=0\right]\right)$

$2.\space \forall x[S(x)=S(y)\to x=y$]

$3.\space x+0=x$

$4.\space x+S(y)=S(x+y)$

$5.\space x\cdot 0=0$

$6.\space x\cdot S(y) = x\cdot y+x$

$7.\space y=0 \vee \exists xS(x)=y$

$\mathbf{Q}$ is modeled by $\mathbf{N}=(\mathbb{N}, 0, S,+,\cdot)$ where $S$ is the successor function and $+, \cdot $ have the usual meaning.

Now take the structure $\mathbf{N}' =\left(\mathbb{N}',0,S', +', \cdot'\right)$ where:

$\mathbb{N}' = \mathbb{N}\cup e$ for some element $e$, $S'(x) = x+S'(0)$ if $x \neq e$ and $e$ otherwise, $x+'e=e$, $x\cdot 'e=e$.

My question is, does it look like $\mathbf{N}'$ models $\mathbf{Q}$? I feel like my modification doesn't allow $e\cdot' 0=0$, which seems to violate axiom 5. I feel like other axioms are satisfied, though.

Answer 1

You haven't defined $+'$ and $\cdot'$ completely. You told use that $x+'e=e$ and $x\cdot 'e=e$, but you have not yet defined $e+'x= ...?$ or $e +' e = ...?$, and technically not even something like $1+'1= ...?$ (though I assume you intend $+'$ and $\cdot'$ to behave like the normal $+$ and $\cdot$ for the natural numbers in the domain .. but you need to spell all of that out)

Now, I assume you want $e+'x = e$ for any $x \in \mathbb{N}'$? and $e + e= e$?

And likewise your thought is: $e\cdot'x = e$ for any $x \in \mathbb{N}'$? and $e \cdot' e= e$?

But now you are indeed in violation of axiom 5. So could you maybe set $e \cdot' 0 = 0$ and set $e \cdot' x = e$ for any $x \not = 0$?

With this change you will indeed satisfy all axioms:

$1.\space \neg\left(\exists x \left[S\left(x\right)=0\right]\right)$

True: $S'(e)=e\not = 0$ and for any $x \not = e$: $S'(x) = x+1 \not = 0$

$2.\space \forall x[S(x)=S(y)\to x=y$]

True. This is the same as saying that no two different objects have the same successor, which is true for all natural numbers in the domain, and the successor of $e$ ois $e$ itself, which is of course differently from all them.

$3.\space x+0=x$

True, since you defined $x +' 0 = x$ for any $x \not = e$, and $e +' 0 = e$

$4.\space x+S(y)=S(x+y)$

True. For $x$ and $y$ natural numbers, this is about the 'normal' addition and successor functions, so it holds, when $x = e$ then $x + S(y) = e +' ... = e$, and when $y = e$ we get $... +' S'(e) = ...+' e = e$

$5.\space x\cdot 0=0$

True. For $x$ a natural number we defined $x \cdot' 0 = 0$, and we also defined $e \cdot 0 = 0$

$6.\space x\cdot S(y) = x\cdot y+x$

True. This is the annoying one. OK, consider the following cases:

$x$ and $y$ are natural numbers. Then it holds given that in that case $\cdot'$, $+'$, and $S'$ work like the 'normal' multiplication, addition, and successor functions

$y = 0$: $x \cdot' S'(y) = x \cdot' S'(0) = x \cdot' 1 = x$ (for both natural number $x$ and $x = e$). While $x \cdot' y +' x = x \cdot' 0 +' x = 0 +' x = x$ (again, for both natural number $x$ and $x = e$). So we're good.

$x= e$ and $y \not = 0$. Then $x \cdot' S'(y) = e \cdot' S'(y) = e$ (since $S'(y) \not = 0$). While $x \cdot' y +' x = ... +' e = e$.

$y = e$: $x \cdot' S'(y) = x \cdot' S'(e) = x \cdot' e = e$. While $x \cdot' y +' x = x \cdot' e +' x = e +' x = e$

This covers all possibilities for $x$ and $y$

$7.\space y=0 \vee \exists x S(x)=y$

True. Indeed every non-zero object in the domain has a predecessor.