Terminology in logic theory

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Edit: I should point out that, I tried searching by myself, but without success. Every time I thought I got the terminology correct, I tried to use it, but, I didn't get relevant results..



I'm taking a logic course in my native language, and I'm having trouble finding the same terminology in english. So here are a few questions.



the first expression, can be assigned True or False, is it called an Atom in english too?



$$ p_i $$
what would an expression like this be called?
$$
p_i land (neg p_j)
$$



what is the term for z called?
$$
z ⊨alpha
$$
here's a question from my homework:



prove the K "isn't definable".
$$
K = Ass backslash z
$$



what I mean mean by "isn't definable", is that there isn't a set like,
$$
Sigma = alpha,beta,...,theta
$$
that:
$$
M(Sigma)=K
$$



I hope I explained myself good enough. And thanks for anyone that would help :)










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  • What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
    – Mauro ALLEGRANZA
    Sep 10 at 11:09






  • 1




    After a few more searches, I think it's called valuation, but I'm not sure...
    – Newone12
    Sep 12 at 8:53














up vote
1
down vote

favorite












Edit: I should point out that, I tried searching by myself, but without success. Every time I thought I got the terminology correct, I tried to use it, but, I didn't get relevant results..



I'm taking a logic course in my native language, and I'm having trouble finding the same terminology in english. So here are a few questions.



the first expression, can be assigned True or False, is it called an Atom in english too?



$$ p_i $$
what would an expression like this be called?
$$
p_i land (neg p_j)
$$



what is the term for z called?
$$
z ⊨alpha
$$
here's a question from my homework:



prove the K "isn't definable".
$$
K = Ass backslash z
$$



what I mean mean by "isn't definable", is that there isn't a set like,
$$
Sigma = alpha,beta,...,theta
$$
that:
$$
M(Sigma)=K
$$



I hope I explained myself good enough. And thanks for anyone that would help :)










share|cite|improve this question























  • What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
    – Mauro ALLEGRANZA
    Sep 10 at 11:09






  • 1




    After a few more searches, I think it's called valuation, but I'm not sure...
    – Newone12
    Sep 12 at 8:53












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Edit: I should point out that, I tried searching by myself, but without success. Every time I thought I got the terminology correct, I tried to use it, but, I didn't get relevant results..



I'm taking a logic course in my native language, and I'm having trouble finding the same terminology in english. So here are a few questions.



the first expression, can be assigned True or False, is it called an Atom in english too?



$$ p_i $$
what would an expression like this be called?
$$
p_i land (neg p_j)
$$



what is the term for z called?
$$
z ⊨alpha
$$
here's a question from my homework:



prove the K "isn't definable".
$$
K = Ass backslash z
$$



what I mean mean by "isn't definable", is that there isn't a set like,
$$
Sigma = alpha,beta,...,theta
$$
that:
$$
M(Sigma)=K
$$



I hope I explained myself good enough. And thanks for anyone that would help :)










share|cite|improve this question















Edit: I should point out that, I tried searching by myself, but without success. Every time I thought I got the terminology correct, I tried to use it, but, I didn't get relevant results..



I'm taking a logic course in my native language, and I'm having trouble finding the same terminology in english. So here are a few questions.



the first expression, can be assigned True or False, is it called an Atom in english too?



$$ p_i $$
what would an expression like this be called?
$$
p_i land (neg p_j)
$$



what is the term for z called?
$$
z ⊨alpha
$$
here's a question from my homework:



prove the K "isn't definable".
$$
K = Ass backslash z
$$



what I mean mean by "isn't definable", is that there isn't a set like,
$$
Sigma = alpha,beta,...,theta
$$
that:
$$
M(Sigma)=K
$$



I hope I explained myself good enough. And thanks for anyone that would help :)







logic propositional-calculus






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edited Sep 10 at 18:49









Mauro ALLEGRANZA

61.5k446105




61.5k446105










asked Sep 10 at 10:39









Newone12

112




112











  • What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
    – Mauro ALLEGRANZA
    Sep 10 at 11:09






  • 1




    After a few more searches, I think it's called valuation, but I'm not sure...
    – Newone12
    Sep 12 at 8:53
















  • What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
    – Mauro ALLEGRANZA
    Sep 10 at 11:09






  • 1




    After a few more searches, I think it's called valuation, but I'm not sure...
    – Newone12
    Sep 12 at 8:53















What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
– Mauro ALLEGRANZA
Sep 10 at 11:09




What is $text Ass$ ? It must be a set of formulas and thus $K = text Ass setminus z $ is the set of formulas obtained from $text Ass$ after "deleting" formula $z$.
– Mauro ALLEGRANZA
Sep 10 at 11:09




1




1




After a few more searches, I think it's called valuation, but I'm not sure...
– Newone12
Sep 12 at 8:53




After a few more searches, I think it's called valuation, but I'm not sure...
– Newone12
Sep 12 at 8:53










1 Answer
1






active

oldest

votes

















up vote
3
down vote













1) Yes, $p_i$ is a propositional variable or atom (or atomic formula): in the context of proposiational logic it represent an "undecomposable" sentence.



2) $p_i ∧ (¬p_j)$ is a formula and specifically a conjunction.



3) In an expression like $Gamma vDash alpha$, meaning that formula $alpha$ is logical consequence of the set $Gamma$ of formulas, we may call the formulas in $Gamma$ premises or assumptions.



But we can have also : $v vDash alpha$, where $v$ is a valuation (also called truth assignment), i.e. an assignment of truth values to propositional variables.



A valuation can be uniquely extended to an assignment of truth values to all propositional formulas.



Thus, $v vDash alpha$ means that valuation $v$ satisfies formula $alpha$, i.e. that $v$ assigns to $alpha$ the value TRUE.



I assume that $text Ass$ is the set of all valuations.



Thus, the question about definability of $K = text Ass setminus z $ amounts to :




find a set $Sigma = p_i mid ldots $ of propositional variables, such that $K$ is the set of models of $Sigma$ (call it : $text Mod (Sigma)$), i.e. the set of valuations that assign to the formulas in $Sigma$ the value TRUE (i.e. $textMod(Sigma) = v mid v text is a valuation and v vDash sigma text for every sigma in Sigma $).







share|cite|improve this answer






















  • Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
    – Newone12
    Sep 12 at 8:13










  • @Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
    – Mauro ALLEGRANZA
    Sep 12 at 8:19










  • I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
    – Newone12
    Sep 12 at 8:50











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













1) Yes, $p_i$ is a propositional variable or atom (or atomic formula): in the context of proposiational logic it represent an "undecomposable" sentence.



2) $p_i ∧ (¬p_j)$ is a formula and specifically a conjunction.



3) In an expression like $Gamma vDash alpha$, meaning that formula $alpha$ is logical consequence of the set $Gamma$ of formulas, we may call the formulas in $Gamma$ premises or assumptions.



But we can have also : $v vDash alpha$, where $v$ is a valuation (also called truth assignment), i.e. an assignment of truth values to propositional variables.



A valuation can be uniquely extended to an assignment of truth values to all propositional formulas.



Thus, $v vDash alpha$ means that valuation $v$ satisfies formula $alpha$, i.e. that $v$ assigns to $alpha$ the value TRUE.



I assume that $text Ass$ is the set of all valuations.



Thus, the question about definability of $K = text Ass setminus z $ amounts to :




find a set $Sigma = p_i mid ldots $ of propositional variables, such that $K$ is the set of models of $Sigma$ (call it : $text Mod (Sigma)$), i.e. the set of valuations that assign to the formulas in $Sigma$ the value TRUE (i.e. $textMod(Sigma) = v mid v text is a valuation and v vDash sigma text for every sigma in Sigma $).







share|cite|improve this answer






















  • Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
    – Newone12
    Sep 12 at 8:13










  • @Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
    – Mauro ALLEGRANZA
    Sep 12 at 8:19










  • I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
    – Newone12
    Sep 12 at 8:50















up vote
3
down vote













1) Yes, $p_i$ is a propositional variable or atom (or atomic formula): in the context of proposiational logic it represent an "undecomposable" sentence.



2) $p_i ∧ (¬p_j)$ is a formula and specifically a conjunction.



3) In an expression like $Gamma vDash alpha$, meaning that formula $alpha$ is logical consequence of the set $Gamma$ of formulas, we may call the formulas in $Gamma$ premises or assumptions.



But we can have also : $v vDash alpha$, where $v$ is a valuation (also called truth assignment), i.e. an assignment of truth values to propositional variables.



A valuation can be uniquely extended to an assignment of truth values to all propositional formulas.



Thus, $v vDash alpha$ means that valuation $v$ satisfies formula $alpha$, i.e. that $v$ assigns to $alpha$ the value TRUE.



I assume that $text Ass$ is the set of all valuations.



Thus, the question about definability of $K = text Ass setminus z $ amounts to :




find a set $Sigma = p_i mid ldots $ of propositional variables, such that $K$ is the set of models of $Sigma$ (call it : $text Mod (Sigma)$), i.e. the set of valuations that assign to the formulas in $Sigma$ the value TRUE (i.e. $textMod(Sigma) = v mid v text is a valuation and v vDash sigma text for every sigma in Sigma $).







share|cite|improve this answer






















  • Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
    – Newone12
    Sep 12 at 8:13










  • @Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
    – Mauro ALLEGRANZA
    Sep 12 at 8:19










  • I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
    – Newone12
    Sep 12 at 8:50













up vote
3
down vote










up vote
3
down vote









1) Yes, $p_i$ is a propositional variable or atom (or atomic formula): in the context of proposiational logic it represent an "undecomposable" sentence.



2) $p_i ∧ (¬p_j)$ is a formula and specifically a conjunction.



3) In an expression like $Gamma vDash alpha$, meaning that formula $alpha$ is logical consequence of the set $Gamma$ of formulas, we may call the formulas in $Gamma$ premises or assumptions.



But we can have also : $v vDash alpha$, where $v$ is a valuation (also called truth assignment), i.e. an assignment of truth values to propositional variables.



A valuation can be uniquely extended to an assignment of truth values to all propositional formulas.



Thus, $v vDash alpha$ means that valuation $v$ satisfies formula $alpha$, i.e. that $v$ assigns to $alpha$ the value TRUE.



I assume that $text Ass$ is the set of all valuations.



Thus, the question about definability of $K = text Ass setminus z $ amounts to :




find a set $Sigma = p_i mid ldots $ of propositional variables, such that $K$ is the set of models of $Sigma$ (call it : $text Mod (Sigma)$), i.e. the set of valuations that assign to the formulas in $Sigma$ the value TRUE (i.e. $textMod(Sigma) = v mid v text is a valuation and v vDash sigma text for every sigma in Sigma $).







share|cite|improve this answer














1) Yes, $p_i$ is a propositional variable or atom (or atomic formula): in the context of proposiational logic it represent an "undecomposable" sentence.



2) $p_i ∧ (¬p_j)$ is a formula and specifically a conjunction.



3) In an expression like $Gamma vDash alpha$, meaning that formula $alpha$ is logical consequence of the set $Gamma$ of formulas, we may call the formulas in $Gamma$ premises or assumptions.



But we can have also : $v vDash alpha$, where $v$ is a valuation (also called truth assignment), i.e. an assignment of truth values to propositional variables.



A valuation can be uniquely extended to an assignment of truth values to all propositional formulas.



Thus, $v vDash alpha$ means that valuation $v$ satisfies formula $alpha$, i.e. that $v$ assigns to $alpha$ the value TRUE.



I assume that $text Ass$ is the set of all valuations.



Thus, the question about definability of $K = text Ass setminus z $ amounts to :




find a set $Sigma = p_i mid ldots $ of propositional variables, such that $K$ is the set of models of $Sigma$ (call it : $text Mod (Sigma)$), i.e. the set of valuations that assign to the formulas in $Sigma$ the value TRUE (i.e. $textMod(Sigma) = v mid v text is a valuation and v vDash sigma text for every sigma in Sigma $).








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 12 at 14:45

























answered Sep 10 at 11:08









Mauro ALLEGRANZA

61.5k446105




61.5k446105











  • Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
    – Newone12
    Sep 12 at 8:13










  • @Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
    – Mauro ALLEGRANZA
    Sep 12 at 8:19










  • I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
    – Newone12
    Sep 12 at 8:50

















  • Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
    – Newone12
    Sep 12 at 8:13










  • @Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
    – Mauro ALLEGRANZA
    Sep 12 at 8:19










  • I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
    – Newone12
    Sep 12 at 8:50
















Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
– Newone12
Sep 12 at 8:13




Thanks for the answer, another small question, is there another way to say "isn't definable", or more correct way to say this?
– Newone12
Sep 12 at 8:13












@Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
– Mauro ALLEGRANZA
Sep 12 at 8:19




@Newone12 - it is not clear what "it is definable" means in the context of propositional calculus. Can you give us some more details, please ?
– Mauro ALLEGRANZA
Sep 12 at 8:19












I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
– Newone12
Sep 12 at 8:50





I'll try to give an example for a "definable" set: for each valuation z, the group: $$ K= z $$ is definable by the group: $$ Sigma = z(p_i)=True bigcup neg p_i $$ and the notation for this would be: $$ M(Sigma)=K $$
– Newone12
Sep 12 at 8:50


















 

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