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If I have an assertion: for all $x$, $x$ equals $5$ Is there any state that satisfies this assertion?

Does the state $x=5$ satisfy the above?

I believe for every $x$, $x$ cannot be $5$. So no state satisfies this assertion?

Stahl
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user93870
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2 Answers2

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What you are calling a state seems to be what it usually called a model. It is a set of elements and relationships between them that satisfy a set of axioms. If you have an axiom $\forall x (x=5)$ the model $\{5\}$ satisfies that axiom. If you have an axiom (your belief) that $\forall x (x \neq 5)$ any model that does not include $5$ satisfies it.

Ross Millikan
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  • Thank you for the answer, one clarification: if the model is set of all integers, there is no way the axiom is satisfied, right? – user93870 Sep 09 '13 at 04:21
  • That is correct, if the model is the set of all integers neither of these two axioms is satisfied. – Ross Millikan Sep 09 '13 at 04:28
  • Thank you...you have been of great help!! – user93870 Sep 09 '13 at 04:29
  • An empty model would also seem to qualify. – Marc van Leeuwen Sep 09 '13 at 04:36
  • @MarcvanLeeuwen: Good point. – Ross Millikan Sep 09 '13 at 13:00
  • @RossMillikan In first order logic, $5$ is a constant, and thus a term. In a standard semantics for first-order logic, the interpretation function maps every term to an element of the domain. If $\forall x,(x \ne 5)$ is a sentence, then $5$ must be mapped to some element of the domain. Another perspective, since $5$ is a term, from $\forall x(x \ne 5)$ and existential introduction, we may infer $\exists y \forall x (x \ne y)$, which cannot be true. Free logic doesn't make this requirement, though. – Joshua Taylor Sep 12 '13 at 20:00
  • @MarcvanLeeuwen Non-empty models typically aren't permitted in first-order logic, as it makes a number of expected inferences fail. – Joshua Taylor Sep 12 '13 at 20:11
  • There's actually a discussion of some of these topics (requiring non-empty domains, interpretations, &c.) in the comments on one of my past answers, as well as the tail end of this answer. – Joshua Taylor Sep 12 '13 at 20:12
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Typically, a model for first-order logic specifies a non-empty domain of discourse $D$ (or universe) and an interpretation function $\cal I$ that maps each constant symbol to an element of the domain, each relation symbol to a relation on the domain (e.g., a 2-ary relation symbol is mapped to a subset of $D \times D$), and each function symbol to a function on $D$.

If a model satisfies the sentence $\forall x (x = 5)$, then what must be true? For every element $e$ of the domain, then $(e,\cal I(5)) \in \cal I(=)$, that is, whatever the interpretation of $5$ in the domain is, every other element of the domain must be the same as it (assuming that $=$ is, in fact, mapped to the actual equivalence relation). Then the domain must be $\{5\}$.

If you let the $\cal I$ map $=$ to something other than the actual equivalence relation on $D$, that is, if you let it be an arbitrary 2-ary relation, you can also come up with some models, but that's not particularly interesting, since you're just saying $\forall x P(x,5)$ for some relation $P$.

Joshua Taylor
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