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I'm a little confused about satisfaction of formulas (in particular of sentences).

Consider a concrete example. Let $L=(\cdot, ^{-1},e)$ be the language of groups and consider its model $M$ with universe $C_3=\{1,g,g^2\}$ and standard interpretations. Let $\phi$ be the formula in the empty set of variables $1\cdot g=g$. Satisfaction is defined inductively. But what case are we dealing with? $g$ is not a term, is it? So $1\cdot g=g$ is not an atomic formula $t_1=t_2$, is it? Also, it's not T or F, nor is it the conjunction/disjunction of atomic formulas... (I didn't mention all cases, but the formula is also not of one of the types I didn't mention, as far as I can see).

user557
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2 Answers2

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Either $g$ is a constant symbol and you should add it to your language. In this case $g$ is a term since symbols of constants are terms. Thus $1.g=g$ is an atomic formula. Or $g$ is a variable, and your formula doesn't have an empty ser of variables.

Régis
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  • I thought $g$ is not a constant symbol. The only constant symbol in the language is $e$ as far as I understand. – user557 Oct 23 '18 at 22:16
  • In this case $g$ is a variable. I edited my answer. – Régis Oct 23 '18 at 22:19
  • I guess my mistake was that I didn't realize that formulas are made up of elements of the language and variables, and I cannot make formulas from the universe of a model. If I did, that would be an interpretation. – user557 Oct 23 '18 at 22:22
  • @Régis: $g$ in this example lives in the metatheory (as would $99$ or $\pi$ in similar examples). – Rob Arthan Oct 23 '18 at 22:37
  • That's a possible way of seeing things. Probably not the less confusing. Anyway the above mentionned formula is not a formula of the language unless $g$ is a symbol of variable (distinct from the object $g$ of the model). – Régis Oct 23 '18 at 22:40
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In general a description of a model takes place in the metalanguage that you are using to talk about the theory whose models you are interested in. In the given description of the model $\{1, g, g^2\}$ of a group, you should read $g$ as a constant in the metatheory. This model is isomorphic to the model with elements $0, 1, 2 \in \Bbb{N}$ and group operation given by addition modulo $3$. Under this isomorphism $1$, $g$ and $g^2$ correspond to $0$, $1$ and $2$ respectively.

Rob Arthan
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  • If you view $1.g=g$ as a formula of the metalanguage, you should mention that then it does not need an interpretation, since interpretation is for formulas of the language. – Régis Oct 23 '18 at 22:48