Re: [Haskell-beginners] Simple function comparison

All right, a revised version: Let f x = x + 1 let g x = x + 2 f ( g a ) = f ( a + 2 ) ................... (g) = ( a + 2 ) + 1 ................. (f) ... = a + 3 ......................... (+) therefore f ( g a ) <-> a + 3 ... (transitivity) g ( f a ) = g ( a + 1 ) ................... (f) = ( a + 1 ) + 2 ................. (g) ... = a + 3 ......................... (+) therefore g ( f a ) <-> a + 3 ... (transitivity) therefore f ( g a ) = g ( f a ) . (transitivity) On 26/11/2015, MJ Williams wrote:

All right, how about this for a proof:

Let f x = x + 1 let g x = x + 2

if f ( g a ) then f ( a + 2 ) .............. (g) then ( a + 2 ) + 1 ............ (f) ... then a + 3 .................... (+) therefore f ( g a ) <-> a + 3

if g ( f a ) then g ( a + 1 ) .............. (f) then ( a + 1 ) + 2 ............ (g) ... then a + 3 .................... (+) therefore g ( f a ) <-> a + 3

therefore f ( g a ) = g ( f a ) (transitivity)

Feel free to take it apart and smash it to bits. Seriously, any feedback welcome.

Cheers, Matthew

On 25/11/2015, Henk-Jan van Tuyl wrote:

...

On Wed, 25 Nov 2015 12:39:15 +0100, Stephen Renehan wrote:

...

Hi,

I'm looking to do a comparison between 2 simple functions to see if they are equivalent but I appear to be running into some problems if anyone can help.

The two functions I want to compare are f (g a) and g (f a).

I have f defined and g defined as f :: a -> a f = undefined

g :: a -> a. g = undefined

As you can not say anything about the type of input or output (except that

they are equal), the only normally terminating function possible is 'id'. So, if you don't want to use 'undefined' or 'error', f and g are equal.

Regards, Henk-Jan van Tuyl

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