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Is `x >> pure y` equivalent to `liftM (const y) x`


Unlike a Functor, a Monad can change shape?Why should Applicative be a superclass of Monad?Is there a monad that doesn't have a corresponding monad transformer (except IO)?Composition of compositions in HaskellHaskell: Flaw in the description of applicative functor laws in the hackage Control.Applicative article?: it says Applicative determines FunctorTo what extent are Applicative/Monad instances uniquely determined?Is this property of a functor stronger than a monad?Are applicative functors composed with the applicative style really independent?bind can be composed of fmap and join, so do we have to use monadic functions a -> m b?Do the monadic liftM and the functorial fmap have to be equivalent?






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15















The two expressions



y >> pure x
liftM (const x) y


have the same type signature in Haskell.
I was curious whether they were equivalent, but I could neither produce a proof of the fact nor a counter example against it.



If we rewrite the two expressions so that we can eliminate the x and y then the question becomes whether the two following functions are equivalent



flip (>>) . pure
liftM . const


Note that both these functions have type Monad m => a -> m b -> m a.



I used the laws that Haskell gives for monad, applicatives, and functors to transform both statements into various equivalent forms, but I was not able to produce a sequence of equivalences between the two.



For instance I found that y >> pure x can be rewritten as follows



y >>= const (pure x)
y *> pure x
(id <$ y) <*> pure x
fmap (const id) y <*> pure x


and liftM (const x) y can be rewritten as follows



fmap (const x) y
pure (const x) <*> y


None of these spring out to me as necessarily equivalent, but I cannot think of any cases where they would not be equivalent.










share|improve this question






























    15















    The two expressions



    y >> pure x
    liftM (const x) y


    have the same type signature in Haskell.
    I was curious whether they were equivalent, but I could neither produce a proof of the fact nor a counter example against it.



    If we rewrite the two expressions so that we can eliminate the x and y then the question becomes whether the two following functions are equivalent



    flip (>>) . pure
    liftM . const


    Note that both these functions have type Monad m => a -> m b -> m a.



    I used the laws that Haskell gives for monad, applicatives, and functors to transform both statements into various equivalent forms, but I was not able to produce a sequence of equivalences between the two.



    For instance I found that y >> pure x can be rewritten as follows



    y >>= const (pure x)
    y *> pure x
    (id <$ y) <*> pure x
    fmap (const id) y <*> pure x


    and liftM (const x) y can be rewritten as follows



    fmap (const x) y
    pure (const x) <*> y


    None of these spring out to me as necessarily equivalent, but I cannot think of any cases where they would not be equivalent.










    share|improve this question


























      15












      15








      15








      The two expressions



      y >> pure x
      liftM (const x) y


      have the same type signature in Haskell.
      I was curious whether they were equivalent, but I could neither produce a proof of the fact nor a counter example against it.



      If we rewrite the two expressions so that we can eliminate the x and y then the question becomes whether the two following functions are equivalent



      flip (>>) . pure
      liftM . const


      Note that both these functions have type Monad m => a -> m b -> m a.



      I used the laws that Haskell gives for monad, applicatives, and functors to transform both statements into various equivalent forms, but I was not able to produce a sequence of equivalences between the two.



      For instance I found that y >> pure x can be rewritten as follows



      y >>= const (pure x)
      y *> pure x
      (id <$ y) <*> pure x
      fmap (const id) y <*> pure x


      and liftM (const x) y can be rewritten as follows



      fmap (const x) y
      pure (const x) <*> y


      None of these spring out to me as necessarily equivalent, but I cannot think of any cases where they would not be equivalent.










      share|improve this question
















      The two expressions



      y >> pure x
      liftM (const x) y


      have the same type signature in Haskell.
      I was curious whether they were equivalent, but I could neither produce a proof of the fact nor a counter example against it.



      If we rewrite the two expressions so that we can eliminate the x and y then the question becomes whether the two following functions are equivalent



      flip (>>) . pure
      liftM . const


      Note that both these functions have type Monad m => a -> m b -> m a.



      I used the laws that Haskell gives for monad, applicatives, and functors to transform both statements into various equivalent forms, but I was not able to produce a sequence of equivalences between the two.



      For instance I found that y >> pure x can be rewritten as follows



      y >>= const (pure x)
      y *> pure x
      (id <$ y) <*> pure x
      fmap (const id) y <*> pure x


      and liftM (const x) y can be rewritten as follows



      fmap (const x) y
      pure (const x) <*> y


      None of these spring out to me as necessarily equivalent, but I cannot think of any cases where they would not be equivalent.







      haskell monads functor applicative






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 27 at 20:30









      duplode

      24.1k45092




      24.1k45092










      asked Mar 27 at 18:31









      10000000001000000000

      489214




      489214






















          3 Answers
          3






          active

          oldest

          votes


















          15














          The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= _ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:



          liftM (const x) y
          = definition of liftM*
          y >>= z -> pure (const x z)
          = definition of const
          y >>= z -> pure x
          = monad law
          y >> pure x


          * Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.






          share|improve this answer

























          • Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

            – chi
            Mar 27 at 20:00






          • 1





            @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

            – Daniel Wagner
            Mar 27 at 20:47







          • 5





            That law itself follows from parametricity and the monad law m >>= pure = m.

            – dfeuer
            Mar 27 at 21:27


















          12














          Yes they are the same



          Let's start with flip (>>) . pure, which is the pointfree version of x >> pure y you provide:



          flip (>>) . pure


          It is the case that flip (>>) is just (=<<) . const so we can rewrite this as:



          ((=<<) . const) . pure


          Since function composition ((.)) is associative we can write this as:



          (=<<) . (const . pure)


          Now we would like to rewrite const . pure. We can notice that const is just pure on (a ->), that means since pure . pure is fmap pure . pure, const . pure is (.) pure . const, ((.) is fmap for the functor (a ->)).



          (=<<) . ((.) pure . const)


          Now we associate again:



          ((=<<) . (.) pure) . const


          ((=<<) . (.) pure) is the definition for liftM1 so we can substitute:



          liftM . const


          And that is the goal. The two are the same.




          1: The definition of liftM is liftM f m1 = do x1 <- m1; return (f x1) , we can desugar the do into liftM f m1 = m1 >>= return . f. We can flip the (>>=) for liftM f m1 = return . f =<< m1 and elide the m1 to get liftM f = (return . f =<<) a little pointfree magic and we get liftM = (=<<) . (.) return






          share|improve this answer




















          • 1





            Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

            – Bergi
            Mar 27 at 22:30







          • 1





            @Bergi Actually you are right, doing it earlier makes things simpler.

            – Sriotchilism O'Zaic
            Mar 27 at 22:54


















          4














          One more possible route, exploiting the applicative laws:




          For instance I found that y >> pure x can be rewritten as follows [...]



          fmap (const id) y <*> pure x



          That amounts to...



          fmap (const id) y <*> pure x
          pure ($ x) <*> fmap (const id) y -- interchange law of applicatives
          fmap ($ x) (fmap (const id) y) -- fmap in terms of <*>
          fmap (($ x) . const id) y -- composition law of functors
          fmap (const x) y


          ... which, as you noted, is the same as liftM (const x) y.



          That this route requires only applicative laws and not monad ones reflects how (*>) (another name for (>>)) is an Applicative method.






          share|improve this answer

























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            3 Answers
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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            15














            The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= _ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:



            liftM (const x) y
            = definition of liftM*
            y >>= z -> pure (const x z)
            = definition of const
            y >>= z -> pure x
            = monad law
            y >> pure x


            * Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.






            share|improve this answer

























            • Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

              – chi
              Mar 27 at 20:00






            • 1





              @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

              – Daniel Wagner
              Mar 27 at 20:47







            • 5





              That law itself follows from parametricity and the monad law m >>= pure = m.

              – dfeuer
              Mar 27 at 21:27















            15














            The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= _ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:



            liftM (const x) y
            = definition of liftM*
            y >>= z -> pure (const x z)
            = definition of const
            y >>= z -> pure x
            = monad law
            y >> pure x


            * Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.






            share|improve this answer

























            • Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

              – chi
              Mar 27 at 20:00






            • 1





              @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

              – Daniel Wagner
              Mar 27 at 20:47







            • 5





              That law itself follows from parametricity and the monad law m >>= pure = m.

              – dfeuer
              Mar 27 at 21:27













            15












            15








            15







            The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= _ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:



            liftM (const x) y
            = definition of liftM*
            y >>= z -> pure (const x z)
            = definition of const
            y >>= z -> pure x
            = monad law
            y >> pure x


            * Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.






            share|improve this answer















            The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= _ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:



            liftM (const x) y
            = definition of liftM*
            y >>= z -> pure (const x z)
            = definition of const
            y >>= z -> pure x
            = monad law
            y >> pure x


            * Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 27 at 18:59

























            answered Mar 27 at 18:54









            Daniel WagnerDaniel Wagner

            105k7163289




            105k7163289












            • Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

              – chi
              Mar 27 at 20:00






            • 1





              @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

              – Daniel Wagner
              Mar 27 at 20:47







            • 5





              That law itself follows from parametricity and the monad law m >>= pure = m.

              – dfeuer
              Mar 27 at 21:27

















            • Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

              – chi
              Mar 27 at 20:00






            • 1





              @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

              – Daniel Wagner
              Mar 27 at 20:47







            • 5





              That law itself follows from parametricity and the monad law m >>= pure = m.

              – dfeuer
              Mar 27 at 21:27
















            Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

            – chi
            Mar 27 at 20:00





            Interesting. For some reason I thought that the standard definition was liftM = fmap, with the more restrictive type. With the real definition above, the wanted equation is much simpler to obtain :)

            – chi
            Mar 27 at 20:00




            1




            1





            @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

            – Daniel Wagner
            Mar 27 at 20:47






            @chi Even without it things aren't too bad: fmap f m = m >>= return . f is also a monad law (one of the oft-forgotten ones).

            – Daniel Wagner
            Mar 27 at 20:47





            5




            5





            That law itself follows from parametricity and the monad law m >>= pure = m.

            – dfeuer
            Mar 27 at 21:27





            That law itself follows from parametricity and the monad law m >>= pure = m.

            – dfeuer
            Mar 27 at 21:27













            12














            Yes they are the same



            Let's start with flip (>>) . pure, which is the pointfree version of x >> pure y you provide:



            flip (>>) . pure


            It is the case that flip (>>) is just (=<<) . const so we can rewrite this as:



            ((=<<) . const) . pure


            Since function composition ((.)) is associative we can write this as:



            (=<<) . (const . pure)


            Now we would like to rewrite const . pure. We can notice that const is just pure on (a ->), that means since pure . pure is fmap pure . pure, const . pure is (.) pure . const, ((.) is fmap for the functor (a ->)).



            (=<<) . ((.) pure . const)


            Now we associate again:



            ((=<<) . (.) pure) . const


            ((=<<) . (.) pure) is the definition for liftM1 so we can substitute:



            liftM . const


            And that is the goal. The two are the same.




            1: The definition of liftM is liftM f m1 = do x1 <- m1; return (f x1) , we can desugar the do into liftM f m1 = m1 >>= return . f. We can flip the (>>=) for liftM f m1 = return . f =<< m1 and elide the m1 to get liftM f = (return . f =<<) a little pointfree magic and we get liftM = (=<<) . (.) return






            share|improve this answer




















            • 1





              Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

              – Bergi
              Mar 27 at 22:30







            • 1





              @Bergi Actually you are right, doing it earlier makes things simpler.

              – Sriotchilism O'Zaic
              Mar 27 at 22:54















            12














            Yes they are the same



            Let's start with flip (>>) . pure, which is the pointfree version of x >> pure y you provide:



            flip (>>) . pure


            It is the case that flip (>>) is just (=<<) . const so we can rewrite this as:



            ((=<<) . const) . pure


            Since function composition ((.)) is associative we can write this as:



            (=<<) . (const . pure)


            Now we would like to rewrite const . pure. We can notice that const is just pure on (a ->), that means since pure . pure is fmap pure . pure, const . pure is (.) pure . const, ((.) is fmap for the functor (a ->)).



            (=<<) . ((.) pure . const)


            Now we associate again:



            ((=<<) . (.) pure) . const


            ((=<<) . (.) pure) is the definition for liftM1 so we can substitute:



            liftM . const


            And that is the goal. The two are the same.




            1: The definition of liftM is liftM f m1 = do x1 <- m1; return (f x1) , we can desugar the do into liftM f m1 = m1 >>= return . f. We can flip the (>>=) for liftM f m1 = return . f =<< m1 and elide the m1 to get liftM f = (return . f =<<) a little pointfree magic and we get liftM = (=<<) . (.) return






            share|improve this answer




















            • 1





              Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

              – Bergi
              Mar 27 at 22:30







            • 1





              @Bergi Actually you are right, doing it earlier makes things simpler.

              – Sriotchilism O'Zaic
              Mar 27 at 22:54













            12












            12








            12







            Yes they are the same



            Let's start with flip (>>) . pure, which is the pointfree version of x >> pure y you provide:



            flip (>>) . pure


            It is the case that flip (>>) is just (=<<) . const so we can rewrite this as:



            ((=<<) . const) . pure


            Since function composition ((.)) is associative we can write this as:



            (=<<) . (const . pure)


            Now we would like to rewrite const . pure. We can notice that const is just pure on (a ->), that means since pure . pure is fmap pure . pure, const . pure is (.) pure . const, ((.) is fmap for the functor (a ->)).



            (=<<) . ((.) pure . const)


            Now we associate again:



            ((=<<) . (.) pure) . const


            ((=<<) . (.) pure) is the definition for liftM1 so we can substitute:



            liftM . const


            And that is the goal. The two are the same.




            1: The definition of liftM is liftM f m1 = do x1 <- m1; return (f x1) , we can desugar the do into liftM f m1 = m1 >>= return . f. We can flip the (>>=) for liftM f m1 = return . f =<< m1 and elide the m1 to get liftM f = (return . f =<<) a little pointfree magic and we get liftM = (=<<) . (.) return






            share|improve this answer















            Yes they are the same



            Let's start with flip (>>) . pure, which is the pointfree version of x >> pure y you provide:



            flip (>>) . pure


            It is the case that flip (>>) is just (=<<) . const so we can rewrite this as:



            ((=<<) . const) . pure


            Since function composition ((.)) is associative we can write this as:



            (=<<) . (const . pure)


            Now we would like to rewrite const . pure. We can notice that const is just pure on (a ->), that means since pure . pure is fmap pure . pure, const . pure is (.) pure . const, ((.) is fmap for the functor (a ->)).



            (=<<) . ((.) pure . const)


            Now we associate again:



            ((=<<) . (.) pure) . const


            ((=<<) . (.) pure) is the definition for liftM1 so we can substitute:



            liftM . const


            And that is the goal. The two are the same.




            1: The definition of liftM is liftM f m1 = do x1 <- m1; return (f x1) , we can desugar the do into liftM f m1 = m1 >>= return . f. We can flip the (>>=) for liftM f m1 = return . f =<< m1 and elide the m1 to get liftM f = (return . f =<<) a little pointfree magic and we get liftM = (=<<) . (.) return







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 27 at 22:54

























            answered Mar 27 at 18:31









            Sriotchilism O'ZaicSriotchilism O'Zaic

            868620




            868620







            • 1





              Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

              – Bergi
              Mar 27 at 22:30







            • 1





              @Bergi Actually you are right, doing it earlier makes things simpler.

              – Sriotchilism O'Zaic
              Mar 27 at 22:54












            • 1





              Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

              – Bergi
              Mar 27 at 22:30







            • 1





              @Bergi Actually you are right, doing it earlier makes things simpler.

              – Sriotchilism O'Zaic
              Mar 27 at 22:54







            1




            1





            Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

            – Bergi
            Mar 27 at 22:30






            Can you please add how you get from const . pure to fmap pure . const? Btw it might have been easier to start with (.) right away instead of writing fmap (and later explaining (figuring out?) what Functor instance it belongs to).

            – Bergi
            Mar 27 at 22:30





            1




            1





            @Bergi Actually you are right, doing it earlier makes things simpler.

            – Sriotchilism O'Zaic
            Mar 27 at 22:54





            @Bergi Actually you are right, doing it earlier makes things simpler.

            – Sriotchilism O'Zaic
            Mar 27 at 22:54











            4














            One more possible route, exploiting the applicative laws:




            For instance I found that y >> pure x can be rewritten as follows [...]



            fmap (const id) y <*> pure x



            That amounts to...



            fmap (const id) y <*> pure x
            pure ($ x) <*> fmap (const id) y -- interchange law of applicatives
            fmap ($ x) (fmap (const id) y) -- fmap in terms of <*>
            fmap (($ x) . const id) y -- composition law of functors
            fmap (const x) y


            ... which, as you noted, is the same as liftM (const x) y.



            That this route requires only applicative laws and not monad ones reflects how (*>) (another name for (>>)) is an Applicative method.






            share|improve this answer





























              4














              One more possible route, exploiting the applicative laws:




              For instance I found that y >> pure x can be rewritten as follows [...]



              fmap (const id) y <*> pure x



              That amounts to...



              fmap (const id) y <*> pure x
              pure ($ x) <*> fmap (const id) y -- interchange law of applicatives
              fmap ($ x) (fmap (const id) y) -- fmap in terms of <*>
              fmap (($ x) . const id) y -- composition law of functors
              fmap (const x) y


              ... which, as you noted, is the same as liftM (const x) y.



              That this route requires only applicative laws and not monad ones reflects how (*>) (another name for (>>)) is an Applicative method.






              share|improve this answer



























                4












                4








                4







                One more possible route, exploiting the applicative laws:




                For instance I found that y >> pure x can be rewritten as follows [...]



                fmap (const id) y <*> pure x



                That amounts to...



                fmap (const id) y <*> pure x
                pure ($ x) <*> fmap (const id) y -- interchange law of applicatives
                fmap ($ x) (fmap (const id) y) -- fmap in terms of <*>
                fmap (($ x) . const id) y -- composition law of functors
                fmap (const x) y


                ... which, as you noted, is the same as liftM (const x) y.



                That this route requires only applicative laws and not monad ones reflects how (*>) (another name for (>>)) is an Applicative method.






                share|improve this answer















                One more possible route, exploiting the applicative laws:




                For instance I found that y >> pure x can be rewritten as follows [...]



                fmap (const id) y <*> pure x



                That amounts to...



                fmap (const id) y <*> pure x
                pure ($ x) <*> fmap (const id) y -- interchange law of applicatives
                fmap ($ x) (fmap (const id) y) -- fmap in terms of <*>
                fmap (($ x) . const id) y -- composition law of functors
                fmap (const x) y


                ... which, as you noted, is the same as liftM (const x) y.



                That this route requires only applicative laws and not monad ones reflects how (*>) (another name for (>>)) is an Applicative method.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 27 at 22:37

























                answered Mar 27 at 20:27









                duplodeduplode

                24.1k45092




                24.1k45092



























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