what does “has a realization” mean?
Consider the following problem:
Suppose $mathcal{M}$ is an $L$-structure and $A subseteq M$ is non-empty. Prove that if every $L_A$-formula in one free variable that has a realisation in $mathcal{M}$ actually has a realisation in $A$, then $A$ is the universe of an elementary substructure of $mathcal{M}$.
What does "has a realisation" mean in this context?
logic model-theory
asked Nov 25 at 0:03
bbw
47038
|
show 1 more comment
Consider the following problem:
Suppose $mathcal{M}$ is an $L$-structure and $A subseteq M$ is non-empty. Prove that if every $L_A$-formula in one free variable that has a realisation in $mathcal{M}$ actually has a realisation in $A$, then $A$ is the universe of an elementary substructure of $mathcal{M}$.
What does "has a realisation" mean in this context?
logic model-theory
asked Nov 25 at 0:03
bbw
47038
1
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
1
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
1
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05
|
show 1 more comment
Consider the following problem:
Suppose $mathcal{M}$ is an $L$-structure and $A subseteq M$ is non-empty. Prove that if every $L_A$-formula in one free variable that has a realisation in $mathcal{M}$ actually has a realisation in $A$, then $A$ is the universe of an elementary substructure of $mathcal{M}$.
What does "has a realisation" mean in this context?
logic model-theory
asked Nov 25 at 0:03
bbw
47038
Consider the following problem:
Suppose $mathcal{M}$ is an $L$-structure and $A subseteq M$ is non-empty. Prove that if every $L_A$-formula in one free variable that has a realisation in $mathcal{M}$ actually has a realisation in $A$, then $A$ is the universe of an elementary substructure of $mathcal{M}$.
What does "has a realisation" mean in this context?
logic model-theory
logic model-theory
asked Nov 25 at 0:03
bbw
47038
asked Nov 25 at 0:03
bbw
47038
asked Nov 25 at 0:03
bbw
47038
asked Nov 25 at 0:03
bbw
47038
asked Nov 25 at 0:03
bbw
47038
47038
1
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
1
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
1
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05
|
show 1 more comment
1
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
1
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
1
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05
1
1
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
1
1
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
1
1
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05
|
show 1 more comment
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1
It means that there is at least one object that will satisfy the formula.
– Bram28
Nov 25 at 0:20
@Bram28, do you mean: "Let $phi(v)$ be an $L_A$-formula in one free variable. If there is some $m in M$ such that $mathcal{M }models phi(m)$, then there will be some $a in A$ such that $mathcal{M} models phi(a)$."?
– bbw
Nov 25 at 0:29
basically yes ... but you have to be careful ... objects from the domain are not the same as constants from yourlanguage
– Bram28
Nov 25 at 0:36
1
@bbw: if there is an $a in M$ with $M vDash phi(a)$ then there is an $a in A$ with $A vDash phi(a)$. The second satisfaction relation is for $A$.
– Carl Mummert
Nov 25 at 1:50
1
@bbw: I would define it in the normal way. There's a lack of context in the problem - what book did it come from? I assume $L$ has no function symbols, so that every subset of a structure is a substructure.
– Carl Mummert
Nov 25 at 4:05