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I was curious whether this statement is satisfied by some model?

And is it satisfied by Herbrand's model?

$∃xR(x)∧¬R(c)$

For the first question, I can take a model where domain is the real numbers and R is whether x is a natural number or not. So this is satisfied by taking x=4 and c=3 for instance. But regarding the second question - if it is not satisfied by some Herbrand model, isn't it suppose to be not satisfied by any model? Or maybe it is indeed satisfied by Herbrand model? But then, what if Herbrand's model is just has only the constant c?

Thanks in advance.

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    If the domain is the irrational numbers, then neither $4$ nor $2$ are there ... – 2017-02-10
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    Oops, I meant saying that the domain is real numbers instead. Sorry. – 2017-02-10
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    Of course, every model of $\exists x R(x) \land \lnot R(c)$ must have at least two elements. – 2017-02-10
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    If the language has only constant $c$ and no function symbol, then the [Herbrand universe](https://en.wikipedia.org/wiki/Herbrand_structure#Definition) will be $U = \{ c \}$. – 2017-02-10
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    Mauro - every model of exists x R(x) and not R(c) must have two constants? Isn't Herbrand universe is only the constant c so only thing I can replace with x is c and then get a contradiction? – 2017-02-10
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    Of course **must** have at least two elements; with only one, call it $c$, the formula will be equiv to $R(c) \land \lnot R(c)$, that is a contradiction. – 2017-02-10

1 Answers 1

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Every model of $∃xR(x) ∧ ¬R(c)$ must have at least two elements.

We can satisfy the formula with e.g. the domain $D = \{ 0, 1 \}$ and interpreting $c$ with $0$ and $R$ with the subset $\{ 1 \}$ of $D$, i.e. with the "property": $(x \ne c)$.

If the language has only the constant $c$ and no function symbol, then the Herbrand universe will be $U = \{ c \}$.

Thus, any Herbrand structure built up on the universe $U$ will not satisfy the formula.

See :

Theorem 9.24 A set of clauses $S$ has a model iff it has an Herbrand model.

Theorem 9.24 is not true if $S$ is an arbitrary formula.

Example 9.25 Let $S = p(a) ∧ ∃x¬p(x)$. Then

$(\{ 0, 1 \}, \{ \{ 0 \} \}, \{ \}, \{ 0 \} )$

is a model for $S$ since $v(p(0)) = T$ , $v(p(1)) = F$.

See page 179 : "An interpretation $\mathfrak I$ is a 4-tuple :

$I = (D, \{ R_1,\ldots, R_l \}, \{ F_1,\ldots, F_m \}, \{ d_1,\ldots, d_n \})$,

where $D$ is a non-empty set called the domain, $R_i$ is an $n_i$-ary relation on $D$ that is assigned to the $n_i$-ary predicate $p_i$, $F_j$ is an $n_j$-ary function on $D$ that is assigned to the function symbol $f_j$ and $d_i ∈ D$ is assigned to the constant $a_i$."

$S$ has no Herbrand models since there are only two Herbrand interpretations and neither is a model:

$( \{ a \}, \{ \{ a \} \}, \{ \}, \{ a \}), (\{ a \}, \{ \{ \} \}, \{ \}, \{ a \})$.

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    I don't get the part where he shows the 4-tuple - what is its use in here? I mean, for that v, why isn't there a Herbrand model? How come there are 2 Herbrand interpretations and neither is a model? How do you receive such result? – 2017-02-10
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    @Eran - how to build the [Herbrand structures](https://en.wikipedia.org/wiki/Herbrand_structure#Definition) ? As said, H universe is $U = \{ c \}$; no function symbols, and thus no possibility to generate more closed terms. The constant symbol $c$ will be interpreted by the element $c \in U$. What about the predicate symol $p$ ? A unary predicate symbol is interpreted with a subset of the domain; the domain $U$ has only two subsets : $\{ \}$ (the emptyset) and $\{ c \}$. Conclusion: only two possible H structures : 1/2 – 2017-02-10
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    $( U , \{ \emptyset \}, \text{no function}, \{ c \} )$ and $( U , \{ U \}, \text{no function}, \{ c \} )$. 2/2 – 2017-02-10
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    Why aren't those Herbrand interpretations models? What I don't get is the second element in the tuple - what is that relation? Why is it a set of sets and how do you compute it? Also - if U={c} is the Herbrand universe. Does not it mean that I can only use c as x and get R(c) and not R(c) and thus the Herbrand model does not satisfy the statement? If so, isn't it known that if an "exists" statement (statements with only "exists" quantifers) is not satisfied for Herbrand, it means that it does not apply to any other model because Herbrand is the smallest model? – 2017-02-10
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    @Eran - sorry, but I'll not copy paste all the book; what defintions (interpretation, Herbrand univers) are you using ? Perhaps, it can be useful if you add them to the text; otherwise, you can like the ref to the relevant pages of your textbook. – 2017-02-10
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    Okay. I get how the structure then. But still I'll ask the more important question of me: Does not it mean that I can only use c as x and get R(c) and not R(c) and thus the Herbrand model does not satisfy the statement? If so, isn't it known that if an "exists" statement (statements with only "exists" quantifers) is not satisfied for Herbrand, it means that it does not apply to any other model because Herbrand is the smallest model? And one more thing - asking about how to say things properly - a Herbrand model exists iff all relations and statements still exists within the new domain, right? – 2017-02-10
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    For a FOL language we need a **domain** $D$ (the first element of the 4-tuple). A way to interpret the *predicate* symbols : for each of them, a suitable relation on $D$. In our case (only one predicate symbol with arity $1$) we need a "unary" relation on $D$, that is a **subset** of $D$ (the second element of the 4-tuple). The same for the *function* symbol (there is none in the example, and thus the third element of the 4-tuple is empty). 1/2 – 2017-02-10
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    Finally, a set of elements of $D$: one for each *constant* symbol. We have only the constant symbol $c$, and thus the set of designata for the constant symbols (the fourth element of the 4-tuple) has only one element. 2/2 – 2017-02-10