Valuations and logic

Question

I have been given the question Let $$Var=\{x,y,z\}$$ $$ U=\{0,1,2\}$$ $$I(p)=\{(0,1),(1,2)\}$$

Give a valuation so that P(x,y) is true?

What is a valuation? I do not understand it intuitively.

Also what does I(p) mean?

Answer 1

Having a first-order language $\mathcal L$ and an interpretation $I$ on a domain (or universe) $U$ of elements, the valuation (or variable assignment) function :

$s : Var \to U$

assigns a "temporary" denotation to the variables.

If we consider the domain $U = \{ 0,1,2 \}$ and a (binary) predicate $P$ such that $I(P) = \{ (0,1),(1,2) \}$, in order to evaluate the truth value of the atomic formula $P(x,y)$, we have to define a valuation $s$.

With an $s$ such that :

$s(x)=0$ and $s(y)=1$,

the formula $P(x,y)$ has now a truth value; this value is "calculated" using the elements of the domain that are the "temporary" denotation of the variables $x,y$ occurring free in the formula.

In our case, we have that $P(x,y)[s]$ is $P(0,1)$, that is true because $(0,1) \in I(P)$.

We write also :

$I,s \vDash P(x,y)$

meaning that the formula $P(x,y)$ is satisfied by the interpretation $I$ with the valuation $s$.

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A slightly more "real" example is obtained for the language of first-order arithmetic.

Let $U$ be the domain $\mathbb N$ of natural numbers with the "usual" interpretation for the (binary) predicate $<$, the (unary) function $S$ ("successor"), the (binary) functions $+$ ("sum") and $\times$ ("product") and the (individual) constant $0$ ("zero").

What is the truth value in this interpretation of the formula :

$x > 0$ ?

If we introduce a valuation $s$ that assign as denotation to the variable $x$ the element $0$ of the domain (i.e. $s(x)=0$), we obtain :

$(x > 0)[s]$ i.e. $0 > 0$, that is false,

while if we use an $s'$ such that $s'(x)=1$, we obtain :

$(x > 0)[s']$ i.e. $1 > 0$, that is true.