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suppose we have a model for a language in first order logic $ M=<D,I> $ such that D is the domain and I is the interpetation such that for every $ a \in D $ we have a closed noun (a noun with no free variables) $ t_0 $ such that $ I[t_0] = a $

prove or disprove the following:

a) if $ \forall x A $ is a sentence than $ M \models \forall x A $ iff for each closed noun $t$ $ M \models A\{t/x\} $ where $ A\{t/x\} $ is substitution of $t$ in place of $x$ in $A$

b) if $ \forall x A $ is a formula than $ M \models \forall x A $ iff for each closed noun $t$ $ M \models A\{t/x\} $

c) if $ \exists x A $ is a sentence than $ M \models \exists x A $ iff there exists a closed noun $t$ such that $ M \models A\{t/x\} $

d) if $ \exists x A $ is a formula than $ M \models \exists x A $ iff there exists a closed noun $t$ such that $ M \models A\{t/x\} $

i think a and c are true because in the case where x is the only free variable in A I can be thought as assignment to $ A\{t/x\} $ (where t is closed noun) but in b and d i have no idea

please help me with this exrecise

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By noun you mean a term? –  tomasz Jun 30 '12 at 19:28
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This question is very context-sensitive. It depends on what definitions of $\models$ you are using. a) and c) sound pretty much like the definition of $\models$ to me. b) also seems to follow by the conventions I know. d) -- not really. Consider $A=(x>y)$ and $M$ a linear order. –  tomasz Jun 30 '12 at 19:36
    
yes noun = term and it isn't the definition the definition is $ M \models A $ if for every assignment v $ M,v \models A $ and $ M \models A\{t/x\} $ means that for every assignment v $ M,v \models A\{t/x\} $ –  bar Jun 30 '12 at 20:33

1 Answer 1

up vote 1 down vote accepted

a) and c) are true, like you said, and I don't think I need to write down the gory details here.

For b) first denote by $B$ the universal closure of $A$ for all variables except $x$. Then we have $M\models \forall x A\iff M\models \forall x B$ (because they're the same thing up to a permutation of universal quantifiers), and the latter is equivalent (by a) ) to that for each closed noun $t$ we have $M\models B\{ t/x\}$, but $t$ is closed, so $B\{t/x\}$ is the universal closure of $A\{t/x\}$, so we have $M\models B\{t/x\}\iff M\models A\{t/x\}$, so b) is true.

For d) we have a counterexample: consider $M=\mathbf Z$ (or any partial order without maximal elements) with $A=(x>y)$. Then we have $M\models \exists x A$, because for any element there is one larger, but for no $k\in Z$ do we have $M\models k>y$, since, for example, we do not have $M\models k>k$.

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