Find an $L$-structure in which the formula is true and an $L$-structure in which the formula is false.

Question

Let $\mathcal{L}=\{P,f\}$, where $P$ is a unary predicate and $f$ is a binary function symbol.

For the $L$-formula find an $\mathcal{L}$-structure $\mathfrak{M}_0$ and a variable assignment $s_0$ over $\mathfrak{M}_0$ in which the formula is true, and find an $\mathcal{L}$-structure $\mathfrak{M}_1$ and a variable assignment $s_1$ over $\mathfrak{M}_1$ in which the formula is false.

$\mathcal{L}$-formula: $\forall v_1f(v_2,v_1)\equiv v_2$

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How am I meant to solve this? It is a longer formula than what I am used to dealing with. I am having a hard time keeping track of everything. I am more used to a simpler $\mathcal{L}$-formula such as:

$Val_{\mathfrak{M},s}(\varphi\wedge\psi)=1$ if $Val_{\mathfrak{M},s}(\varphi)=1$ and $Val_{\mathfrak{M},s}(\varphi)=1$

$Val_{\mathfrak{M},s}(\varphi\wedge\psi)=0$ if $Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\varphi)=0$

Answer 1

For the $\mathcal L$-structure $\mathfrak M_0$, consider the structure with domain the set $\mathbb N$ of natural numbers and interpret the binary function symbol $f$ with the "product" ($\times$) operation.

Then, the $\mathcal L$-formula :

$∀v_1 f(v_2,v_1) = v_2$

means :

$\forall x (y \times x = y)$.

Thus, with the variable assignment $s_0$ such that : $s_0(y)=0$, we have that :

$\mathfrak M_0 \vDash \forall x (f(y,x) = y)[s_0]$

because in $\mathbb N$ the formula : $\forall x (0 \times x = 0)$ holds.

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For the $\mathcal L$-structure $\mathfrak M_1$, consider agian $\mathbb N$ and interpret $f$ as "sum" ($+$).

In this case the formula means :

$\forall x (y + x = y)$.

Now consider $s_1$ such that $s_1(y)=1$, and we have that :

$\mathfrak M_1 \nvDash \forall x (f(y,x) = y)[s_1]$

because in $\mathbb N$ the formula : $\forall x (1 + x = 1)$ does not hold.