What's a set of sentential formulas satisfied by $\aleph_0$ assignments in $\{0,1\}$-logic but $2^{\aleph_0}$ assignments in $\{0,1/3,2/3,1\}$-logic?

Question

$\{0,1/3,2/3,1\}$ logic has the following connectives:

$\neg r= 1-r$

$r\wedge s=min\{r,s\}$

$r\vee s=max\{r,s\}$

$r\rightarrow s=min\{1,1+s-r\}$

$r\oplus s=min\{r+s,2-r-s\}$

$r\leftrightarrow s=min\{1+r-s,1+s-r\}$

Assume there are countably many atoms. All connectives are allowed in constructing the set. But truth constants are not allowed.

I know that sets like $\{A_{k+1}\rightarrow A_k: k \in \mathbb{N}\}$ or $\{A_i \vee A_j: i,j\in \mathbb{N}, i\neq j\}$ have $\aleph_0$ truth assignments satisfying them in $\{0,1\}$ logic, but I cannot think of an example that furthermore has $2^{\aleph_0}$ truth assignments satisfying it in $\{0,1/3,2/3,1\}$ logic. Any hints?

Answer 1

I assume throughout that "satisfies" means "gives truth value $1$" in your four-valued logic.

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Given an atom $A_i$, consider the sentence $$(*)_i:\quad (A_i\vee\neg A_i)\rightarrow A_i.$$ The truth assignments satisfying $(*)_i$ in two-valued logic are exactly those giving $A_i$ truth value $1$, while the truth assignments satisfying $(*)_i$ in your four-valued logic are more complicated. There are two key points:

• Truth assignments giving $A_i$ value $0$ or $1$ give $(A_i\vee \neg A_i)$ value $1$, while truth assignments giving $A_i$ value $1\over 3$ or ${2\over 3}$ give $(A_i\vee \neg A_i)$ value $2\over 3$.

• A truth assignment satisfies an implication $X\rightarrow Y$ iff it gives $Y$ a truth value at least as large as the truth value it gives $X$.

Putting these together, we get a characterization of those truth assignments satisfying $(*)_i$:

They are exactly the truth assignments giving $A_i$ value either ${2\over 3}$ or $1$.

This gives rise to a theory which behaves very differently in the two logics in consideration, namely $$\Theta=\{(*)_i: i\in\mathbb{N}\}.$$ There are continuum-many truth assignments satisfying $\Theta$ in your four-valued logic, but only one truth assignment satisfying it in two-valued logic.

Of course this doesn't answer your question yet since $1\not=\aleph_0$, but it does give an important first step. In particular, by combining $\Theta$ with a theory which has $\aleph_0$-many satisfying truth assignments in each logic you'll get what you want:

Consider for example $$\{(*)_{2i}: i\in\mathbb{N}\}\cup\{A_{2i+3}\rightarrow A_{2i+1}: i\in\mathbb{N}\}.$$