1

From the book Logic for Mathematicians (A. G. Hamilton):

Definition 3.19

Two valuations $v$ and $v'$ are $i$-equivalent if $v(x_j)=v'(x_j)$ for all $j\neq i$.

(I know every single book out there have a different way to define this.)

Then consider $A= (\forall x_1)(\exists x_2) P(x_1,f(x_2))$ under the interpretation $\cal I$ with domain $\mathbb{Z}$, and $P(x_1,x_2)$ means $x_1=x_2$, and $f(x)=x+1$.

It is easy to see that $A$ is true for $\cal I$, but my goal is to write the proof in a more or less formal way (similar to the one that is handled in the same book).

This is what I have now:

Consider $v$ to be a valuation in $\cal I$, then $v(x_1), v(x_2) \in \mathbb{Z}$ and we interpret $P(x_1,f(x_2))$ as $v(x_1)=v(x_2)+1$ which is true for $v(x_1)=2$ and $v(x_2)=1$. Now we start to consider $i$-equivalent valuations. So for every valuation $v'$ 1-equivalent to $v$, there always exists a valuation $v'$ 2-equivalent to $v$ in which $v'(x_1)=v'(x_2)+1$.

I don't know if that's OK, and really I don't know if I'm understanding valuations well at all.

Community
  • 1
Wyvern666
  • 911

1 Answers1

2

I think you've grasped the concept of valuations. As you've noted, not every presentation of interpretations for first-order logic does this exactly the same way. The Wikipedia article on interpretations in logic has a section interpretations of a first-order language, and you'll notice that they first describe the easy-to-grasp intuition of how to understand whether an interpretation satisfies $\forall x.\phi(x)$ and $\exists x.\phi(x)$.

This leaves the issue of how to interpret formulas of the form ∀ x φ(x) and ∃ x φ(x). The domain of discourse forms the range for these quantifiers. The idea is that the sentence ∀ x φ(x) is true under an interpretation exactly when every substitution instance of φ(x), where x is replaced by some element of the domain, is satisfied. The formula ∃ x φ(x) is satisfied if there is at least one element d of the domain such that φ(d) is satisfied. — Interpretations of a first-order language

The problem arises however, that we can't actually talk about $\phi(d)$, because that isn't actually a formula:

Strictly speaking, a substitution instance such as the formula φ(d) mentioned above is not a formula in the original formal language of φ, because d is an element of the domain. There are two ways of handling this technical issue. The first is to pass to a larger language in which each element of the domain is named by a constant symbol. The second is to add to the interpretation a function that assigns each variable to an element of the domain. Then the T-schema can quantify over variations of the original interpretation in which this variable assignment function is changed, instead of quantifying over substitution instances. — Interpretations of a first-order language

The definition given by Hamilton follows the second of these options. It does mean, however, that we no longer get to speak of an interpretation $\cal I$ satisfying a formula, but we have to speak of $\cal I$ and a valuation $v$ satisfying a formula. The alternative is to include a valuation $v$ inside an interpretation, but we still end up having to say that $\cal I = (D,\dots,v)$ (where $\dots$ is everything else that is in the interpretation) satisfies $\forall x.\phi(x)$ if and only if every $\cal I' = (D,\dots,v')$ satisfies $\forall x.\phi(x)$, where $v'$ differs from $v'$ only on $x$. This amounts to the same thing, but with more syntax.

Your proof is in pretty good shape as it is, I think. You could express some things a bit more rigorously, perhaps, but the intuition and general structure is correct. Since you're defining the domain to be $\mathbb{Z}$ and can reference mathematical functions in your semantic explanation, I think you can do something like the following. Since you are interpreting $P$ as the equality function on $\mathbb{Z}$ and $f$ as $+1$, I simply used equality in the following formula, and $s$ as the successor ($+1$) function.

Interpretation $\cal I$ and valuation $v$ satisfy $\forall x_1.\exists x_2.(x_1 = s(x_2))$ if and only if $\cal I$ and every 1-equivalent valuation $v'$ satisfy $\exists x_2.(x_1 = s(x_2))$. The interpretation $\cal I$ and valuation $v'$ satisfy $\exists x_2.(x_1 = s(x_2))$ if and only if there is some valuation $v''$ that is 2-equivalent to $v'$ such that $\cal I$ and $v''$ satisfy $x_1 = s(x_2)$.

Consider an arbitrary valuation $v$. Let $v'$ be an arbitary valuation that is 1-equivalent to $v$. Then let $n = v'(x_1)$. Let $v''$ be the valuation that agrees with $v'$ on every value except $x_2$, which it maps to $n-1$. Observe that $\cal I$ and $v''$ satisfy $x_1 = s(x_2)$. Then, since there is some valuation (viz., $v''$) that is 2-equivalent to $v'$ which, with $\cal I$, satisfies $x_1 = s(x_2)$, then $\cal I$ and $v'$ satisfy $\exists x_2.(x_1 = s(x_2))$. Then, since $v'$ is an arbitrary valuation 1-equivalent to $v$, every valuation 1-equivalent to $v$, with $\cal I$, satisfies $\exists x_2.(x_1 = s(x_2))$, so $\cal I$ and $v$ satisfy $\forall x_1.\exists x_2.(x_1 = s(x_2))$.

Community
  • 1
Joshua Taylor
  • 2,481
  • 16
  • 35
  • There have been some other questions and answers that have included discussion of first order interpretations. The last few paragraphs of this answer about the meaning of symbols (e.g., unbound variables) in Fitch-style proofs discuss interpretations, as does a section in this answer about proving sentences based on their semantics. – Joshua Taylor Jun 14 '13 at 04:05
  • I only used $v$ and $v'$, you added $v''$. Well, i think i understand the proof, the concept about it, but what is more difficult to understand is the "syntax" used. Let me correct mi proof... when i consider $v'$ 1-equivalent to $v$, then if i want to consider another situacion inside this $v'$ (in this case about $x_2$) i need to refer to $v''$ 2-equivalent $v'$ ??. – Wyvern666 Jun 14 '13 at 04:31
  • Another question, ¿is really needed to say: "$v′′$ be the valuation that agrees with $v′$ on every value except $x_2$" ?, i tought that was clarified when you say "$v''$ 2-equivalent to $v'$" so the only value to be changed from $v'$ is $x_2$. – Wyvern666 Jun 14 '13 at 05:00
  • 1
    @Wyvern666 To first comment: Suppose we ask whether interpretation $\cal I$ and valuation $v$ satisfy formula $\forall x.\exists y.\phi(x,y)$. Then we first consider whether every $v'$ that is $x$-equivalent to $v$ satisfies $\exists y.\phi(y)$. For each $v'$, we ask whether $\cal I, v'$ satisfy $\exists y.\phi(x,y)$; i.e., whether there is some $v''$ that is $y$-equivalent to $v'$ such that $\cal I, v''$ satisfy $\phi(x,y)$. All the $v'$ give us all the possible valuations for $x$, and for each one, $v''$ provides the particular value of $y$. – Joshua Taylor Jun 14 '13 at 11:00
  • 1
    @Wyvern666 To second comment: I wrote “let $v''$ be the valuation that agrees…” because I wanted to highlight that we can construct the particular $v''$ that we need, and then observe that is 2-equivalent to $v'$ (which we need for the semantics of $\exists x_2.\phi(x)$). It's probably OK to just say, “let $v''$ be the value that is 2-equivalent to $v'$ and that maps $x_2$ to $n-1$.” – Joshua Taylor Jun 14 '13 at 11:10
  • Now im getting it. This means that if i have more vars suppose $\forall x,\exists y,\exists z$, then is probable that i will end up working with $v,v',v'',v'''$ valuations in order to make a valid proof. So depending on the case i will have just one or maybe n valuations to consider? – Wyvern666 Jun 14 '13 at 15:26
  • 1
    @Wyvern666 Well, if you try to keep up this level of rigor, you will. In practice it's more common (and easier) to just say, “$\forall x.\exists y.(x = y+1)$ is true because $x$ must be some value $n$ from $\mathbb{Z}$, and for each such $n$, $n-1$ is a suitable $y$. It's interesting how informal the formal logicians are. :) – Joshua Taylor Jun 14 '13 at 15:30
  • Your explanations are excelent. Thanks. – Wyvern666 Jun 14 '13 at 15:34
  • More on this topic, when dealing with negated cuantifiers, ¿is more convenient to express them in some equivalent not negated form in order to make the proof?. – Wyvern666 Jun 19 '13 at 16:28
  • 1
    @Wyvern666 I think it very much depends on what the end goal is. $\exists$ and $\forall$ are duals, as are $\lor$ and $\land$. We can ask whether it's more convenient to have $\lnot(p \lor q)$ or $\lnot p \land \lnot q$, but the answer depends on what we're trying to prove. – Joshua Taylor Jun 19 '13 at 16:35
  • Yes i understand what you mean. But, for example, if you have $\lnot \forall x P(x)$, so i need to find every valuation x-equivalent that NOT satisfies P(x)?. I think is more intuitive to find what satisfies and not the other way.. Maybe i just need more practice. – Wyvern666 Jun 19 '13 at 17:07