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Connection between logic and set theory?


Axioms of set theory and logicResolution Proof QuestionWhy $¬¬bot notin PROP$?Set theory and logic proofHow can I prove this relation between the elementary set theory and the elementary logic?What is the connection between game theory and (modal) logic?Problem on elementary logic and set theoryDifferences between set theory and second order logic?Logic and set theory (De Morgan's law)Logic and set theory - proofs of theorems













8












$begingroup$


I just noticed there is a similarity between logic operations on propositions and the operations of set theory. It seems:



$$beginarrayllll
textrmdisjunction & (-)vee (-)& textrmcorresponds to union& (-)cup (-)\
textrmconjunction & (-) wedge (-)& textrmcorrespons to intersection & (-)cap (-)\
textrmnegation & sim (-) & textrmcorrespons to taking complements & c(-),
endarray
$$

and I conjecture:



$$beginarraylll
textrmconditional & (-)rightarrow (-) & textrmcorresponds to inclusion & subset\
textrmbiconditional & (-)leftrightarrow (-) & textrmcorresponds to equality & =
endarray$$



How far does it go? I believe there is some kind of functor between some category whose objects are propositions and the category of sets, is that right?



Thanks










share|cite|improve this question











$endgroup$











  • $begingroup$
    The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
    $endgroup$
    – GEdgar
    Mar 21 at 12:57






  • 1




    $begingroup$
    For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:11











  • $begingroup$
    See also The Algebra of Logic Tradition for historical development.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:12










  • $begingroup$
    Thanks for the answers!
    $endgroup$
    – PtF
    Mar 21 at 13:53






  • 2




    $begingroup$
    The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
    $endgroup$
    – Henning Makholm
    Mar 21 at 15:54
















8












$begingroup$


I just noticed there is a similarity between logic operations on propositions and the operations of set theory. It seems:



$$beginarrayllll
textrmdisjunction & (-)vee (-)& textrmcorresponds to union& (-)cup (-)\
textrmconjunction & (-) wedge (-)& textrmcorrespons to intersection & (-)cap (-)\
textrmnegation & sim (-) & textrmcorrespons to taking complements & c(-),
endarray
$$

and I conjecture:



$$beginarraylll
textrmconditional & (-)rightarrow (-) & textrmcorresponds to inclusion & subset\
textrmbiconditional & (-)leftrightarrow (-) & textrmcorresponds to equality & =
endarray$$



How far does it go? I believe there is some kind of functor between some category whose objects are propositions and the category of sets, is that right?



Thanks










share|cite|improve this question











$endgroup$











  • $begingroup$
    The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
    $endgroup$
    – GEdgar
    Mar 21 at 12:57






  • 1




    $begingroup$
    For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:11











  • $begingroup$
    See also The Algebra of Logic Tradition for historical development.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:12










  • $begingroup$
    Thanks for the answers!
    $endgroup$
    – PtF
    Mar 21 at 13:53






  • 2




    $begingroup$
    The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
    $endgroup$
    – Henning Makholm
    Mar 21 at 15:54














8












8








8


1



$begingroup$


I just noticed there is a similarity between logic operations on propositions and the operations of set theory. It seems:



$$beginarrayllll
textrmdisjunction & (-)vee (-)& textrmcorresponds to union& (-)cup (-)\
textrmconjunction & (-) wedge (-)& textrmcorrespons to intersection & (-)cap (-)\
textrmnegation & sim (-) & textrmcorrespons to taking complements & c(-),
endarray
$$

and I conjecture:



$$beginarraylll
textrmconditional & (-)rightarrow (-) & textrmcorresponds to inclusion & subset\
textrmbiconditional & (-)leftrightarrow (-) & textrmcorresponds to equality & =
endarray$$



How far does it go? I believe there is some kind of functor between some category whose objects are propositions and the category of sets, is that right?



Thanks










share|cite|improve this question











$endgroup$




I just noticed there is a similarity between logic operations on propositions and the operations of set theory. It seems:



$$beginarrayllll
textrmdisjunction & (-)vee (-)& textrmcorresponds to union& (-)cup (-)\
textrmconjunction & (-) wedge (-)& textrmcorrespons to intersection & (-)cap (-)\
textrmnegation & sim (-) & textrmcorrespons to taking complements & c(-),
endarray
$$

and I conjecture:



$$beginarraylll
textrmconditional & (-)rightarrow (-) & textrmcorresponds to inclusion & subset\
textrmbiconditional & (-)leftrightarrow (-) & textrmcorresponds to equality & =
endarray$$



How far does it go? I believe there is some kind of functor between some category whose objects are propositions and the category of sets, is that right?



Thanks







elementary-set-theory logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 15:48









Andrés E. Caicedo

65.8k8160252




65.8k8160252










asked Mar 21 at 12:47









PtFPtF

4,10821735




4,10821735











  • $begingroup$
    The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
    $endgroup$
    – GEdgar
    Mar 21 at 12:57






  • 1




    $begingroup$
    For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:11











  • $begingroup$
    See also The Algebra of Logic Tradition for historical development.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:12










  • $begingroup$
    Thanks for the answers!
    $endgroup$
    – PtF
    Mar 21 at 13:53






  • 2




    $begingroup$
    The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
    $endgroup$
    – Henning Makholm
    Mar 21 at 15:54

















  • $begingroup$
    The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
    $endgroup$
    – GEdgar
    Mar 21 at 12:57






  • 1




    $begingroup$
    For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:11











  • $begingroup$
    See also The Algebra of Logic Tradition for historical development.
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 21 at 13:12










  • $begingroup$
    Thanks for the answers!
    $endgroup$
    – PtF
    Mar 21 at 13:53






  • 2




    $begingroup$
    The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
    $endgroup$
    – Henning Makholm
    Mar 21 at 15:54
















$begingroup$
The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
$endgroup$
– GEdgar
Mar 21 at 12:57




$begingroup$
The category of sets deals with functions from one set to another. What will be the arrows in your proposed category where the objects are propositions? Summary: Yes, there is a parallel here, but you should not bring in categories.
$endgroup$
– GEdgar
Mar 21 at 12:57




1




1




$begingroup$
For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 13:11





$begingroup$
For conenctio, see Boolean algebra as well as The Mathematics of Boolean Algebra.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 13:11













$begingroup$
See also The Algebra of Logic Tradition for historical development.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 13:12




$begingroup$
See also The Algebra of Logic Tradition for historical development.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 13:12












$begingroup$
Thanks for the answers!
$endgroup$
– PtF
Mar 21 at 13:53




$begingroup$
Thanks for the answers!
$endgroup$
– PtF
Mar 21 at 13:53




2




2




$begingroup$
The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
$endgroup$
– Henning Makholm
Mar 21 at 15:54





$begingroup$
The parallel between $to$ and $subseteq$ is not quite direct. Note that $to$ combines two propositions into a proposition that you can use with further logical symbols, whereas $subseteq$ combines two set expressions into a proposition. So whereas you can say $Aland(Bto C)$ in logic, you cannot meaningfully say $Acap(Bsubseteq C)$ in set algebra. (Similarly for $leftrightarrow$ versus $=$).
$endgroup$
– Henning Makholm
Mar 21 at 15:54











3 Answers
3






active

oldest

votes


















4












$begingroup$

"How far does it go ?" : as far as it can get !



Here's a general idea about how set theory is a semantics for classical propositional logic (note that if we change the formal system we're looking at without changing our assumptions on sets, for instance if we're studying intuitionistic logic from a classical point of view, then we have to take another semantics, in this specific case, topological spaces can be appropriate) :



Suppose you have a set of propositional variables $V$, a "global" set $E$, and a function $[-] :Vto mathcalP(E)$. Then you can build a function that goes from the set $mathrmForm$ of formulas to $mathcalP(E)$ by expanding $[-]$ according to the rules you displayed : if $varphi, psi$ are formulas and you already defined $[varphi]$ and $[psi]$, then define $[varphi land psi] = [varphi]cap [psi]$, similarly for $lor, neg$, and define $[varphiimplies psi]$ as $xin Emid xin [varphi]implies xin[psi]$.



These rules allow you to define $[varphi]$ for any formula $varphi$ by induction, going from the variables and gaining complexity.



Then you can prove the following things : if $varphi$ is a theorem of classical logic, then $[varphi] = E$, which tells you that the set-operations behave according to the logical ones, but also you can prove : if for any $E$ and any $[-] :Vto mathcalP(E)$, $[varphi] = E$, then $varphi$ is a theorem of classical logic ! This tells you that actually the logical operations behave just like set-theoretic operations as well.



There's actually a lot more you can say about this sort of thing (for instance : what happens if you add quantifiers ? Or in another direction what happens if we replace $mathcalP(E)$ by some other type of object ? If we completely change the type of object, what kind of logic do we get ? etc. etc.)



If you absolutely want to use the words "functor" and "category" you can, but at this level they're not the most relevant thing.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Yes, there is an abstract isomorphism here and the likeness of the symbols $lor$ and $cup$, as well as that of $land$ and $cap$ is of course no accident!



    Also, if you use the formal definition of the set operators, we see the connection there as well:



    Union:



    $forall A forall B forall x (x in A colorredcup B leftrightarrow (x in A colorredlor x in B))$



    Intersection:



    $forall A forall B forall x (x in A colorredcap B leftrightarrow (x in A colorredland x in B))$



    Complement:



    $forall A forall B forall x (x in Acolorred' leftrightarrow colorredneg x in A)$



    And your conjecture is right in that we also have:



    Inclusion:



    $forall A forall B (Acolorred subseteq B leftrightarrow forall x (x in A colorred rightarrow x in B))$



    Equality:



    $forall A forall B (Acolorred = B leftrightarrow forall x (x in A colorred leftrightarrow x in B))$






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      The connection is math is a logical science. It's built on a foundation of axioms and definitions, from which ideally, all statements can be proven, or disproven ( though some are undecideable sadly). You can have logical properties with set operations, Sets equipped with operations, having certain properties applied on the set, are the basis for: magmas, monoids, loops, semigroups, quasigroups, groups, abelian groups, rings, commutative rings, and fields, just to name a few buzzwords. Mathematical logic, can be defined in terms of other more basic logics.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        "How far does it go ?" : as far as it can get !



        Here's a general idea about how set theory is a semantics for classical propositional logic (note that if we change the formal system we're looking at without changing our assumptions on sets, for instance if we're studying intuitionistic logic from a classical point of view, then we have to take another semantics, in this specific case, topological spaces can be appropriate) :



        Suppose you have a set of propositional variables $V$, a "global" set $E$, and a function $[-] :Vto mathcalP(E)$. Then you can build a function that goes from the set $mathrmForm$ of formulas to $mathcalP(E)$ by expanding $[-]$ according to the rules you displayed : if $varphi, psi$ are formulas and you already defined $[varphi]$ and $[psi]$, then define $[varphi land psi] = [varphi]cap [psi]$, similarly for $lor, neg$, and define $[varphiimplies psi]$ as $xin Emid xin [varphi]implies xin[psi]$.



        These rules allow you to define $[varphi]$ for any formula $varphi$ by induction, going from the variables and gaining complexity.



        Then you can prove the following things : if $varphi$ is a theorem of classical logic, then $[varphi] = E$, which tells you that the set-operations behave according to the logical ones, but also you can prove : if for any $E$ and any $[-] :Vto mathcalP(E)$, $[varphi] = E$, then $varphi$ is a theorem of classical logic ! This tells you that actually the logical operations behave just like set-theoretic operations as well.



        There's actually a lot more you can say about this sort of thing (for instance : what happens if you add quantifiers ? Or in another direction what happens if we replace $mathcalP(E)$ by some other type of object ? If we completely change the type of object, what kind of logic do we get ? etc. etc.)



        If you absolutely want to use the words "functor" and "category" you can, but at this level they're not the most relevant thing.






        share|cite|improve this answer









        $endgroup$

















          4












          $begingroup$

          "How far does it go ?" : as far as it can get !



          Here's a general idea about how set theory is a semantics for classical propositional logic (note that if we change the formal system we're looking at without changing our assumptions on sets, for instance if we're studying intuitionistic logic from a classical point of view, then we have to take another semantics, in this specific case, topological spaces can be appropriate) :



          Suppose you have a set of propositional variables $V$, a "global" set $E$, and a function $[-] :Vto mathcalP(E)$. Then you can build a function that goes from the set $mathrmForm$ of formulas to $mathcalP(E)$ by expanding $[-]$ according to the rules you displayed : if $varphi, psi$ are formulas and you already defined $[varphi]$ and $[psi]$, then define $[varphi land psi] = [varphi]cap [psi]$, similarly for $lor, neg$, and define $[varphiimplies psi]$ as $xin Emid xin [varphi]implies xin[psi]$.



          These rules allow you to define $[varphi]$ for any formula $varphi$ by induction, going from the variables and gaining complexity.



          Then you can prove the following things : if $varphi$ is a theorem of classical logic, then $[varphi] = E$, which tells you that the set-operations behave according to the logical ones, but also you can prove : if for any $E$ and any $[-] :Vto mathcalP(E)$, $[varphi] = E$, then $varphi$ is a theorem of classical logic ! This tells you that actually the logical operations behave just like set-theoretic operations as well.



          There's actually a lot more you can say about this sort of thing (for instance : what happens if you add quantifiers ? Or in another direction what happens if we replace $mathcalP(E)$ by some other type of object ? If we completely change the type of object, what kind of logic do we get ? etc. etc.)



          If you absolutely want to use the words "functor" and "category" you can, but at this level they're not the most relevant thing.






          share|cite|improve this answer









          $endgroup$















            4












            4








            4





            $begingroup$

            "How far does it go ?" : as far as it can get !



            Here's a general idea about how set theory is a semantics for classical propositional logic (note that if we change the formal system we're looking at without changing our assumptions on sets, for instance if we're studying intuitionistic logic from a classical point of view, then we have to take another semantics, in this specific case, topological spaces can be appropriate) :



            Suppose you have a set of propositional variables $V$, a "global" set $E$, and a function $[-] :Vto mathcalP(E)$. Then you can build a function that goes from the set $mathrmForm$ of formulas to $mathcalP(E)$ by expanding $[-]$ according to the rules you displayed : if $varphi, psi$ are formulas and you already defined $[varphi]$ and $[psi]$, then define $[varphi land psi] = [varphi]cap [psi]$, similarly for $lor, neg$, and define $[varphiimplies psi]$ as $xin Emid xin [varphi]implies xin[psi]$.



            These rules allow you to define $[varphi]$ for any formula $varphi$ by induction, going from the variables and gaining complexity.



            Then you can prove the following things : if $varphi$ is a theorem of classical logic, then $[varphi] = E$, which tells you that the set-operations behave according to the logical ones, but also you can prove : if for any $E$ and any $[-] :Vto mathcalP(E)$, $[varphi] = E$, then $varphi$ is a theorem of classical logic ! This tells you that actually the logical operations behave just like set-theoretic operations as well.



            There's actually a lot more you can say about this sort of thing (for instance : what happens if you add quantifiers ? Or in another direction what happens if we replace $mathcalP(E)$ by some other type of object ? If we completely change the type of object, what kind of logic do we get ? etc. etc.)



            If you absolutely want to use the words "functor" and "category" you can, but at this level they're not the most relevant thing.






            share|cite|improve this answer









            $endgroup$



            "How far does it go ?" : as far as it can get !



            Here's a general idea about how set theory is a semantics for classical propositional logic (note that if we change the formal system we're looking at without changing our assumptions on sets, for instance if we're studying intuitionistic logic from a classical point of view, then we have to take another semantics, in this specific case, topological spaces can be appropriate) :



            Suppose you have a set of propositional variables $V$, a "global" set $E$, and a function $[-] :Vto mathcalP(E)$. Then you can build a function that goes from the set $mathrmForm$ of formulas to $mathcalP(E)$ by expanding $[-]$ according to the rules you displayed : if $varphi, psi$ are formulas and you already defined $[varphi]$ and $[psi]$, then define $[varphi land psi] = [varphi]cap [psi]$, similarly for $lor, neg$, and define $[varphiimplies psi]$ as $xin Emid xin [varphi]implies xin[psi]$.



            These rules allow you to define $[varphi]$ for any formula $varphi$ by induction, going from the variables and gaining complexity.



            Then you can prove the following things : if $varphi$ is a theorem of classical logic, then $[varphi] = E$, which tells you that the set-operations behave according to the logical ones, but also you can prove : if for any $E$ and any $[-] :Vto mathcalP(E)$, $[varphi] = E$, then $varphi$ is a theorem of classical logic ! This tells you that actually the logical operations behave just like set-theoretic operations as well.



            There's actually a lot more you can say about this sort of thing (for instance : what happens if you add quantifiers ? Or in another direction what happens if we replace $mathcalP(E)$ by some other type of object ? If we completely change the type of object, what kind of logic do we get ? etc. etc.)



            If you absolutely want to use the words "functor" and "category" you can, but at this level they're not the most relevant thing.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 21 at 19:38









            MaxMax

            15.9k11144




            15.9k11144





















                3












                $begingroup$

                Yes, there is an abstract isomorphism here and the likeness of the symbols $lor$ and $cup$, as well as that of $land$ and $cap$ is of course no accident!



                Also, if you use the formal definition of the set operators, we see the connection there as well:



                Union:



                $forall A forall B forall x (x in A colorredcup B leftrightarrow (x in A colorredlor x in B))$



                Intersection:



                $forall A forall B forall x (x in A colorredcap B leftrightarrow (x in A colorredland x in B))$



                Complement:



                $forall A forall B forall x (x in Acolorred' leftrightarrow colorredneg x in A)$



                And your conjecture is right in that we also have:



                Inclusion:



                $forall A forall B (Acolorred subseteq B leftrightarrow forall x (x in A colorred rightarrow x in B))$



                Equality:



                $forall A forall B (Acolorred = B leftrightarrow forall x (x in A colorred leftrightarrow x in B))$






                share|cite|improve this answer









                $endgroup$

















                  3












                  $begingroup$

                  Yes, there is an abstract isomorphism here and the likeness of the symbols $lor$ and $cup$, as well as that of $land$ and $cap$ is of course no accident!



                  Also, if you use the formal definition of the set operators, we see the connection there as well:



                  Union:



                  $forall A forall B forall x (x in A colorredcup B leftrightarrow (x in A colorredlor x in B))$



                  Intersection:



                  $forall A forall B forall x (x in A colorredcap B leftrightarrow (x in A colorredland x in B))$



                  Complement:



                  $forall A forall B forall x (x in Acolorred' leftrightarrow colorredneg x in A)$



                  And your conjecture is right in that we also have:



                  Inclusion:



                  $forall A forall B (Acolorred subseteq B leftrightarrow forall x (x in A colorred rightarrow x in B))$



                  Equality:



                  $forall A forall B (Acolorred = B leftrightarrow forall x (x in A colorred leftrightarrow x in B))$






                  share|cite|improve this answer









                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    Yes, there is an abstract isomorphism here and the likeness of the symbols $lor$ and $cup$, as well as that of $land$ and $cap$ is of course no accident!



                    Also, if you use the formal definition of the set operators, we see the connection there as well:



                    Union:



                    $forall A forall B forall x (x in A colorredcup B leftrightarrow (x in A colorredlor x in B))$



                    Intersection:



                    $forall A forall B forall x (x in A colorredcap B leftrightarrow (x in A colorredland x in B))$



                    Complement:



                    $forall A forall B forall x (x in Acolorred' leftrightarrow colorredneg x in A)$



                    And your conjecture is right in that we also have:



                    Inclusion:



                    $forall A forall B (Acolorred subseteq B leftrightarrow forall x (x in A colorred rightarrow x in B))$



                    Equality:



                    $forall A forall B (Acolorred = B leftrightarrow forall x (x in A colorred leftrightarrow x in B))$






                    share|cite|improve this answer









                    $endgroup$



                    Yes, there is an abstract isomorphism here and the likeness of the symbols $lor$ and $cup$, as well as that of $land$ and $cap$ is of course no accident!



                    Also, if you use the formal definition of the set operators, we see the connection there as well:



                    Union:



                    $forall A forall B forall x (x in A colorredcup B leftrightarrow (x in A colorredlor x in B))$



                    Intersection:



                    $forall A forall B forall x (x in A colorredcap B leftrightarrow (x in A colorredland x in B))$



                    Complement:



                    $forall A forall B forall x (x in Acolorred' leftrightarrow colorredneg x in A)$



                    And your conjecture is right in that we also have:



                    Inclusion:



                    $forall A forall B (Acolorred subseteq B leftrightarrow forall x (x in A colorred rightarrow x in B))$



                    Equality:



                    $forall A forall B (Acolorred = B leftrightarrow forall x (x in A colorred leftrightarrow x in B))$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 21 at 15:00









                    Bram28Bram28

                    64.2k44793




                    64.2k44793





















                        1












                        $begingroup$

                        The connection is math is a logical science. It's built on a foundation of axioms and definitions, from which ideally, all statements can be proven, or disproven ( though some are undecideable sadly). You can have logical properties with set operations, Sets equipped with operations, having certain properties applied on the set, are the basis for: magmas, monoids, loops, semigroups, quasigroups, groups, abelian groups, rings, commutative rings, and fields, just to name a few buzzwords. Mathematical logic, can be defined in terms of other more basic logics.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          The connection is math is a logical science. It's built on a foundation of axioms and definitions, from which ideally, all statements can be proven, or disproven ( though some are undecideable sadly). You can have logical properties with set operations, Sets equipped with operations, having certain properties applied on the set, are the basis for: magmas, monoids, loops, semigroups, quasigroups, groups, abelian groups, rings, commutative rings, and fields, just to name a few buzzwords. Mathematical logic, can be defined in terms of other more basic logics.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            The connection is math is a logical science. It's built on a foundation of axioms and definitions, from which ideally, all statements can be proven, or disproven ( though some are undecideable sadly). You can have logical properties with set operations, Sets equipped with operations, having certain properties applied on the set, are the basis for: magmas, monoids, loops, semigroups, quasigroups, groups, abelian groups, rings, commutative rings, and fields, just to name a few buzzwords. Mathematical logic, can be defined in terms of other more basic logics.






                            share|cite|improve this answer









                            $endgroup$



                            The connection is math is a logical science. It's built on a foundation of axioms and definitions, from which ideally, all statements can be proven, or disproven ( though some are undecideable sadly). You can have logical properties with set operations, Sets equipped with operations, having certain properties applied on the set, are the basis for: magmas, monoids, loops, semigroups, quasigroups, groups, abelian groups, rings, commutative rings, and fields, just to name a few buzzwords. Mathematical logic, can be defined in terms of other more basic logics.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 21 at 20:46









                            Roddy MacPheeRoddy MacPhee

                            648118




                            648118



























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