“Many” quantifier in first order logic [closed]

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Suppose I want to express this sentence in first-order logic:



"Many printers are broken"


If $p(x)$ is true if the term $x$ is a printer and $b(x)$ is true if $x$ is broken, how can I express this sentence?






logic first-order-logic predicate-logic quantifiers logic-translation







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closed as off-topic by Shaun, Adrian Keister, Leucippus, user99914, Xander Henderson Sep 11 at 1:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Leucippus, Community, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
    – Mauro ALLEGRANZA
    Sep 10 at 19:55














up vote
-2
down vote

favorite












Suppose I want to express this sentence in first-order logic:



"Many printers are broken"


If $p(x)$ is true if the term $x$ is a printer and $b(x)$ is true if $x$ is broken, how can I express this sentence?






logic first-order-logic predicate-logic quantifiers logic-translation







share|cite|improve this question















closed as off-topic by Shaun, Adrian Keister, Leucippus, user99914, Xander Henderson Sep 11 at 1:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Leucippus, Community, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
    – Mauro ALLEGRANZA
    Sep 10 at 19:55












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











Suppose I want to express this sentence in first-order logic:



"Many printers are broken"


If $p(x)$ is true if the term $x$ is a printer and $b(x)$ is true if $x$ is broken, how can I express this sentence?






logic first-order-logic predicate-logic quantifiers logic-translation







share|cite|improve this question















Suppose I want to express this sentence in first-order logic:



"Many printers are broken"


If $p(x)$ is true if the term $x$ is a printer and $b(x)$ is true if $x$ is broken, how can I express this sentence?






logic first-order-logic predicate-logic quantifiers logic-translation




logic first-order-logic predicate-logic quantifiers logic-translation






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edited Sep 11 at 13:14









Taroccoesbrocco

4,17461535




4,17461535










asked Sep 10 at 19:50









Koinos

757




757




closed as off-topic by Shaun, Adrian Keister, Leucippus, user99914, Xander Henderson Sep 11 at 1:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Leucippus, Community, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Shaun, Adrian Keister, Leucippus, user99914, Xander Henderson Sep 11 at 1:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Leucippus, Community, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
    – Mauro ALLEGRANZA
    Sep 10 at 19:55












  • 1




    You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
    – Mauro ALLEGRANZA
    Sep 10 at 19:55







1




1




You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
– Mauro ALLEGRANZA
Sep 10 at 19:55




You cannot; you can see Generalized quantifiers : see e.g. $text most_M (A,B)$ ("most $A$ are $B$").
– Mauro ALLEGRANZA
Sep 10 at 19:55










1 Answer
1






active

oldest

votes

















up vote
6
down vote



accepted










First-order logic does not have a notion of "many" - this, like "most," "almost all," etc. is an example of a generalized quantifier, and handling them takes us to extensions of first-order logic.



We can express each of the following in first-order logic:



  • "Some printer is broken." ($exists x(p(x)wedge b(x))$)


  • "Multiple printers are broken." ($exists x,y(xnot=y wedge p(x)wedge p(y)wedge b(x)wedge b(y))$)


  • "At least $n$ printers are broken." ($exists x_1,...,x_n((bigwedge_1le i<jle nx_inot=x_j)wedge (bigwedge_1le kle np(x_k)wedge b(x_k))$)


  • "All printers are broken." ($forall x(p(x)implies b(x))$)






share|cite|improve this answer




















  • Is the inequality between terms an extension of first order logic ?
    – Koinos
    Sep 11 at 9:58










  • @Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
    – Noah Schweber
    Sep 11 at 14:17

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote



accepted










First-order logic does not have a notion of "many" - this, like "most," "almost all," etc. is an example of a generalized quantifier, and handling them takes us to extensions of first-order logic.



We can express each of the following in first-order logic:



  • "Some printer is broken." ($exists x(p(x)wedge b(x))$)


  • "Multiple printers are broken." ($exists x,y(xnot=y wedge p(x)wedge p(y)wedge b(x)wedge b(y))$)


  • "At least $n$ printers are broken." ($exists x_1,...,x_n((bigwedge_1le i<jle nx_inot=x_j)wedge (bigwedge_1le kle np(x_k)wedge b(x_k))$)


  • "All printers are broken." ($forall x(p(x)implies b(x))$)






share|cite|improve this answer




















  • Is the inequality between terms an extension of first order logic ?
    – Koinos
    Sep 11 at 9:58










  • @Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
    – Noah Schweber
    Sep 11 at 14:17














up vote
6
down vote



accepted










First-order logic does not have a notion of "many" - this, like "most," "almost all," etc. is an example of a generalized quantifier, and handling them takes us to extensions of first-order logic.



We can express each of the following in first-order logic:



  • "Some printer is broken." ($exists x(p(x)wedge b(x))$)


  • "Multiple printers are broken." ($exists x,y(xnot=y wedge p(x)wedge p(y)wedge b(x)wedge b(y))$)


  • "At least $n$ printers are broken." ($exists x_1,...,x_n((bigwedge_1le i<jle nx_inot=x_j)wedge (bigwedge_1le kle np(x_k)wedge b(x_k))$)


  • "All printers are broken." ($forall x(p(x)implies b(x))$)






share|cite|improve this answer




















  • Is the inequality between terms an extension of first order logic ?
    – Koinos
    Sep 11 at 9:58










  • @Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
    – Noah Schweber
    Sep 11 at 14:17












up vote
6
down vote



accepted







up vote
6
down vote



accepted






First-order logic does not have a notion of "many" - this, like "most," "almost all," etc. is an example of a generalized quantifier, and handling them takes us to extensions of first-order logic.



We can express each of the following in first-order logic:



  • "Some printer is broken." ($exists x(p(x)wedge b(x))$)


  • "Multiple printers are broken." ($exists x,y(xnot=y wedge p(x)wedge p(y)wedge b(x)wedge b(y))$)


  • "At least $n$ printers are broken." ($exists x_1,...,x_n((bigwedge_1le i<jle nx_inot=x_j)wedge (bigwedge_1le kle np(x_k)wedge b(x_k))$)


  • "All printers are broken." ($forall x(p(x)implies b(x))$)






share|cite|improve this answer












First-order logic does not have a notion of "many" - this, like "most," "almost all," etc. is an example of a generalized quantifier, and handling them takes us to extensions of first-order logic.



We can express each of the following in first-order logic:



  • "Some printer is broken." ($exists x(p(x)wedge b(x))$)


  • "Multiple printers are broken." ($exists x,y(xnot=y wedge p(x)wedge p(y)wedge b(x)wedge b(y))$)


  • "At least $n$ printers are broken." ($exists x_1,...,x_n((bigwedge_1le i<jle nx_inot=x_j)wedge (bigwedge_1le kle np(x_k)wedge b(x_k))$)


  • "All printers are broken." ($forall x(p(x)implies b(x))$)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 10 at 19:55









Noah Schweber

113k9143267




113k9143267











  • Is the inequality between terms an extension of first order logic ?
    – Koinos
    Sep 11 at 9:58










  • @Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
    – Noah Schweber
    Sep 11 at 14:17
















  • Is the inequality between terms an extension of first order logic ?
    – Koinos
    Sep 11 at 9:58










  • @Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
    – Noah Schweber
    Sep 11 at 14:17















Is the inequality between terms an extension of first order logic ?
– Koinos
Sep 11 at 9:58




Is the inequality between terms an extension of first order logic ?
– Koinos
Sep 11 at 9:58












@Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
– Noah Schweber
Sep 11 at 14:17




@Koinos You mean "$not=$?" No - "$=$" is a logical symbol in first-order logic (this is actually historically contingent - early versions of first-order logic didn't always include $=$), and "$anot=b$" is understood as an abbreviation for "$neg(a=b)$ (similarly, "$exists x,y$" is an abbreviation for "$exists xexists y$," and etc.).
– Noah Schweber
Sep 11 at 14:17


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