$mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$ is an algebra












-1












$begingroup$



Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



Show that $mathscr{C}$ is an algebra.




To show $mathscr{C}$ is an algebra:



i) $emptysetinmathscr{C}$



ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



For example if I want to check $mathscr{C}$ is closed for unions then:



If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



Question:



It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



Thanks in advance!






measure-theory proof-verification elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$



    Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



    Show that $mathscr{C}$ is an algebra.




    To show $mathscr{C}$ is an algebra:



    i) $emptysetinmathscr{C}$



    ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



    iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



    iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



    For example if I want to check $mathscr{C}$ is closed for unions then:



    If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



    Question:



    It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



    Thanks in advance!






    measure-theory proof-verification elementary-set-theory proof-writing







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$



      Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



      Show that $mathscr{C}$ is an algebra.




      To show $mathscr{C}$ is an algebra:



      i) $emptysetinmathscr{C}$



      ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



      iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



      iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



      For example if I want to check $mathscr{C}$ is closed for unions then:



      If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



      Question:



      It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



      Thanks in advance!






      measure-theory proof-verification elementary-set-theory proof-writing







      share|cite|improve this question











      $endgroup$





      Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



      Show that $mathscr{C}$ is an algebra.




      To show $mathscr{C}$ is an algebra:



      i) $emptysetinmathscr{C}$



      ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



      iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



      iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



      For example if I want to check $mathscr{C}$ is closed for unions then:



      If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



      Question:



      It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



      Thanks in advance!






      measure-theory proof-verification elementary-set-theory proof-writing




      measure-theory proof-verification elementary-set-theory proof-writing






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 22:05









      Andrés E. Caicedo

      65.4k8158249




      65.4k8158249










      asked Jan 18 at 19:30









      Pedro GomesPedro Gomes

      1,8262721




      1,8262721






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15



















          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23













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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15
















          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15














          1












          1








          1





          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$



          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 20:29

























          answered Jan 18 at 19:36









          greedoidgreedoid

          42.4k1153105




          42.4k1153105












          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15


















          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15
















          $begingroup$
          How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
          $endgroup$
          – Pedro Gomes
          Jan 18 at 19:39




          $begingroup$
          How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
          $endgroup$
          – Pedro Gomes
          Jan 18 at 19:39












          $begingroup$
          I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Jan 18 at 21:23






          $begingroup$
          I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Jan 18 at 21:23














          $begingroup$
          Yes, but is it in $mathscr{C}$?
          $endgroup$
          – greedoid
          Jan 19 at 18:06






          $begingroup$
          Yes, but is it in $mathscr{C}$?
          $endgroup$
          – greedoid
          Jan 19 at 18:06














          $begingroup$
          Check my answer!
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:15




          $begingroup$
          Check my answer!
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:15











          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23


















          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23
















          0












          0








          0





          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$



          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 18:14









          Pedro GomesPedro Gomes

          1,8262721




          1,8262721












          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23




















          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23


















          $begingroup$
          Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
          $endgroup$
          – greedoid
          Jan 19 at 18:20






          $begingroup$
          Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
          $endgroup$
          – greedoid
          Jan 19 at 18:20














          $begingroup$
          Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:23






          $begingroup$
          Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:23




















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          Mario Kart Wii

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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          The Binding of Isaac: Rebirth/Afterbirth

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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