Are my formulas correct and are there any tips and tricks to find them faster?

Question

I have the following Problem.

Be $\sigma=\{G, \cdot\}$ a signature in which $G$ is a digit relations symbol and $\cdot$ two-digit function symbol. Be $\mathcal{A}=\left(\mathbb{Z}, G^{\mathcal{A}},\cdot^{\mathcal{A}}\right)$ a $\sigma$-structure in which $G^{\mathcal{A}}=\{2 x | x \in \mathbb{Z}\}$ and $\cdot^\mathcal{A}$ is the normal integer multiplication.

Give a formula for $\varphi_{1},\varphi_{2}$ and $\varphi_{3}$ so that

$(i) \varphi_{1}(\mathcal{A})=\{1\}$

(ii) $\varphi_{2}(\mathcal{A})=\{-1\}$

(iii) $\varphi_{3}(\mathcal{A})=\left\{s 2^{k} | k \in \mathbb{N} \text { und } s \in\{1,-1\}\right\}$

Now I want to know three things: First, is my solution correct.

$(i) \varphi_{1}(x) = \forall{y}(x \cdot y = y)$

$(ii) \varphi_{2}(x) = \exists{y}(x \cdot x = y \wedge \varphi_{1}(y))$

$(iii) \varphi_{3}(x) = \exists{y} \exists{z}(y * G(z) = x \wedge (\varphi_{1}(y) \vee \varphi_{2}(y)))$

Second, is $(ii) \varphi_{2}(x) = \exists{y}(x \cdot x = \varphi_{1}(y))$ also ok ?

Third, if everything is wrong, what how could I change it, to be correct, are there any tricks?

The alternative formulation of (ii) is not syntactically correct; $\varphi_1$ is a formula and $=$ must be flanked by terms. — Mauro ALLEGRANZA, Feb 19 '20 at 12:01
IMO, (i) and (ii) are correct; I'm not sure about (iii) — Mauro ALLEGRANZA, Feb 19 '20 at 12:03
$G(x)$ is "$x$ is Even" while (iii) must define the set of powers of $2$. In your formula we have that $\mathcal A \vDash \varphi_3(6)$ because $6 \in G^A$ and $6=6 \times 1$ but there is no $k$ such that $6=2^k$ — Mauro ALLEGRANZA, Feb 19 '20 at 12:08
Exponentiation is not easily managed in first-order arithmetic; see How is exponentiation defined in Peano arithmetic? — Mauro ALLEGRANZA, Feb 19 '20 at 12:14
@MauroALLEGRANZA Yes, (iii) is false :( This is a hart task:) — philipa, Feb 19 '20 at 13:08
In (ii) you need to assert $\lnot \phi_1(x)$ so that $\phi_2(1)$ is false. — Rob Arthan, Feb 20 '20 at 21:44

Rob Arthan · Answer 1 · 2022-02-23T21:20:53.257

The comments have dealt with (i) and (ii). For (iii), an integer has the form $\pm 2^k$ iff all its divisors are either units or even:

$$\varphi_3(x) \equiv \forall y\forall z(x = y \cdot z \Rightarrow (\varphi_1(y) \lor \varphi_2(y)\lor G(y)))$$

or, equivalently, iff all its odd divisors are units:

$$\varphi_3(x) \equiv \forall y\forall z((x = y \cdot z \land \lnot G(y)) \Rightarrow (\varphi_1(y) \lor \varphi_2(y)))$$

EDIT [Added for the record much later]

Your proposed solution for (ii) has the main idea but isn't quite right: your $\varphi_2(x)$ is true when $x=1$ as well as when $x = -1$. This is easy to fix (and we can also get rid of the existential quantification, because we only need to talk about $x$ and $x\cdot x$): we make $\phi_2(x)$ assert that $x \cdot x$ is $1$ but $x$ is not $1$:

$$ \varphi_2(x) \equiv \phi_1(x\cdot x) \land \lnot\phi_1(x) $$

Are my formulas correct and are there any tips and tricks to find them faster?

1 Answers1