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Hey I am having a hard time solving a logic problem. It one of the exercises in Ebbinghaus, Mathematical Logic (exercise 4.14, p. 39):

A set $\phi$ of S-formulas is independent if there is no $\varphi \in \phi$ s.t. $\phi $/$ \{\varphi\} \vDash \varphi$. Let $\mathcal{S}=\{\circ, e\}$ and $$\phi := \{ \forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z), \forall x \ x \circ e \equiv x, \forall x \exists y \ x \circ y \equiv e \}\text{.}$$ Show that $\phi$ is independent.

My thoughts: I need to find three models $M1$, $M2$, $M3$ s.t. $M1$ satisfies the second and third formulas, but cannot satisfy the first formula $\forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z)$, so the first formula is independent.

$M2$, $M3$ are similar cases. i,e., $M2$ can satisfy the first formula and the third formula, but cannot satisfy the second formula, so the second one is independent. $M3$ can satisfy the first formula and the second formula, but cannot satisfy the third formula, so the third one is independent.

Like, consider M3, let the domain of M3 is $\mathbb{Z}$, $\varphi(\circ)$ $= +$ on $\mathbb{Z}$，and $\varphi(e) = 0$, so $M3$ satisfies the first formula (associative law) and the second $\mathbb{Z}+0=0$. But $M3$ cannot satisfy the third one, so the third formula is independent.

But I'm wondering :

(1) Could this kind of specific model works? I mean I just find a "specific model" to satisfy two formulas in the set. I just feel I need to prove independence in a more general way, right? But I have no idea how to do.

(2) Could anyone think examples of $M1$ and $M2$?

Any ideas please? I appreciate your help.

Bernard
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Giorgio
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    $(\mathbb Z, +)$ IS a group. So your model doesn't work. It satisfies the third formula because every integer has an additive inverse. – jjagmath Mar 10 '21 at 11:12
  • @jjagmath Ahh okay... so I thought an invalid model. But do you know if this method works? Or do you have any valid model? Thanks! – Giorgio Mar 10 '21 at 11:30
  • Use $\mathbb N = {0,1,2,\cdots }$ instead of $\mathbb Z$ – jjagmath Mar 10 '21 at 11:35
  • To address part (1) of the question: individual examples are all you need to prove independence. Your approach is fine, but you just need to find the right examples. – Rob Arthan Mar 10 '21 at 12:55

2 Answers2

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I'll just write in a very non-rigorous way but you can get some ideas.

For $M_3$, take any monoid that is not a group, such as $(\mathbb N,+)$.

For $M_1$, consider $(\mathbb Z,-)$. More generally, for most groups $(G,\cdot)$, define a new operation $x\circ y=x\cdot y^{-1}$, and use it on $G$. (Won't work so easily if your axioms are two-sided though. Maybe the octonions?)

For $M_2$, take any group but interpret $e$ differently (not identity). For example take $e=1$ in $(\mathbb Z,+)$.

Tesla Daybreak
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For $M_1$ I propose the following minimal possible model. An operation on a set of two elements $\{e,a\}$ is given by the Cayley table $$ \begin{array}{c|cc} \circ & e & a\\ \hline e & e & e \\ a & a & e \\ \end{array} $$ We have $(a\circ a)\circ a=e\circ a=e$ and $a\circ(a\circ a)=a\circ e=a$.

kabenyuk
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