Propositions in Haskell

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Patrick Browne

15 May 2013 15 May '13

11:34 a.m.

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Dan Mead

15 May 15 May

12:34 p.m.

i don't understand what you're trying to do with that code, however you seem to be asking about theorem proving in general check out http://www.haskell.org/haskellwiki/Libraries_and_tools/Theorem_provers and http://en.wikipedia.org/wiki/Automated_theorem_proving hope it helps On Wed, May 15, 2013 at 11:34 AM, Patrick Browne wrote:

...

-- Hi -- I am trying to show that a set of propositions and a conclusion the form a valid argument. -- I used two approaches; 1) using if-then-else, 2) using pattern matching. -- The version using if-then-else seems to be consistent with my knowledge of Haskell and logic (either of which could be wrong). -- Can the second approach be improved to better reflect the propositions and conclusion? Maybe type level reasoning could be used? -- -- Valid argument? -- 1. I work hard or I play piano -- 2. If I work hard then I will get a bonus -- 3. But I did not get a bonus -- Therefore I played piano -- Variables: p = Piano, w = worked hard, b = got a bonus -- (w \/ p) /\ (w => b) /\ ¬(b) -- --------------------------- -- p

-- First approach using language control structure if-then-else w, p, b::Bool -- Two equivalences for (w \/ p) as an implication. -- 1. (w \/ p) =equivalent-to=> (not p) => w -- 2. (w \/ p) =equivalent-to=> (not w) => p -- Picked 2 p = if (not w) then True else False -- Contrapositive: (w => b) =equivalent-to=> ~b => ~w w = if (not b) then False else True b = False -- gives p is true and w is false

-- Second approach using pattern matching -- I think the rewriting goes from left to right but the logical inference goes in the opposite direction. w1, p1, b1::Bool p1 = (not w1) w1 = b1 -- Not consistent with statements, but I do not know how to write ~b1 => ~w1 in Haskell b1 = False -- Again gives p1 is true and w1 is false

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

MigMit

2:04 p.m.

You can stop suspecting: in Haskell, equations ARE definitions. On May 15, 2013, at 9:15 PM, Patrick Browne wrote:

...

The relation to theorem proving is the main motivation for my question.

In am trying to understand why some equations are ok and others not.

I suspect that in Haskell equations are definitions rather than assertions.

If approach 2 is a non-starter in Haskell, then can approach 1, using if-then-else, achieve the same results for propositions?

Thanks Pat

On 15/05/13, Dan Mead wrote:

...
i don't understand what you're trying to do with that code, however you seem to be asking about theorem proving in general

check out

http://www.haskell.org/haskellwiki/Libraries_and_tools/Theorem_provers

and

http://en.wikipedia.org/wiki/Automated_theorem_proving

hope it helps

On Wed, May 15, 2013 at 11:34 AM, Patrick Browne > wrote: -- Hi -- I am trying to show that a set of propositions and a conclusion the form a valid argument. -- I used two approaches; 1) using if-then-else, 2) using pattern matching. -- The version using if-then-else seems to be consistent with my knowledge of Haskell and logic (either of which could be wrong). -- Can the second approach be improved to better reflect the propositions and conclusion? Maybe type level reasoning could be used? -- -- Valid argument? -- 1. I work hard or I play piano -- 2. If I work hard then I will get a bonus -- 3. But I did not get a bonus -- Therefore I played piano -- Variables: p = Piano, w = worked hard, b = got a bonus -- (w \/ p) /\ (w => b) /\ ¬(b) -- --------------------------- -- p

-- First approach using language control structure if-then-else w, p, b::Bool -- Two equivalences for (w \/ p) as an implication. -- 1. (w \/ p) =equivalent-to=> (not p) => w -- 2. (w \/ p) =equivalent-to=> (not w) => p -- Picked 2 p = if (not w) then True else False -- Contrapositive: (w => b) =equivalent-to=> ~b => ~w w = if (not b) then False else True b = False -- gives p is true and w is false

-- Second approach using pattern matching -- I think the rewriting goes from left to right but the logical inference goes in the opposite direction. w1, p1, b1::Bool p1 = (not w1) w1 = b1 -- Not consistent with statements, but I do not know how to write ~b1 => ~w1 in Haskell b1 = False -- Again gives p1 is true and w1 is false

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Tillmann Rendel

17 May 17 May

4:58 p.m.

Hi, Patrick Browne wrote:

...

In am trying to understand why some equations are ok and others not.

I suspect that in Haskell equations are definitions rather than assertions.

Yes. Haskell function definitions look like equations, but in many ways, they aren't. Here is an example for an equation that is not valid as a Haskell function definition: g x = 42 f (g x) = x -- not valid The problem is that we cannot use g at the left-hand side. Here is an example that doesn't mean what you might be hoping for: f x = f x f x = 42 Seen as an equation system, this should constrain f to be a function that always returns 42. But in Haskell, it defines f to be a non-terminating function. The reason is that only the first line counts, because it covers all possible argument values. The second line is ignored. The behavior changes if we swap the two lines: g x = 42 g x = g x Again, only the first line counts, so g is the function that always returns 42. Here is a more complicated example: h 27 = 42 h x = h x h 13 = 100 What function is h? Tillmann

Alberto G. Corona

15 May 15 May

12:56 p.m.

Not exactly what you ask, but it is noteworthy that the mind has different logic processors. The fastest one work with IF THEN ELSE rules applied specifically to deals. This is why your example (and most examples of logic) involves a kind of deal expressed in the first person. This trigger a fast mental evaluation, while an equivalent but more general case is harder to process and need some paper work. (That special treatment of first person deals logic respond to the need to detect breaks of deals as fast as possible) http://en.wikipedia.org/wiki/Wason_selection_task That's why higher level languages have redundant logical structures and do not follow a general abstract and short mathematical notation. Therefore "higher level", in programming languages, does not mean higher mathematical abstraction, but to be closer to the way the mind works. 2013/5/15 Patrick Browne

...

-- Hi -- I am trying to show that a set of propositions and a conclusion the form a valid argument. -- I used two approaches; 1) using if-then-else, 2) using pattern matching. -- The version using if-then-else seems to be consistent with my knowledge of Haskell and logic (either of which could be wrong). -- Can the second approach be improved to better reflect the propositions and conclusion? Maybe type level reasoning could be used? -- -- Valid argument? -- 1. I work hard or I play piano -- 2. If I work hard then I will get a bonus -- 3. But I did not get a bonus -- Therefore I played piano -- Variables: p = Piano, w = worked hard, b = got a bonus -- (w \/ p) /\ (w => b) /\ ¬(b) -- --------------------------- -- p

-- First approach using language control structure if-then-else w, p, b::Bool -- Two equivalences for (w \/ p) as an implication. -- 1. (w \/ p) =equivalent-to=> (not p) => w -- 2. (w \/ p) =equivalent-to=> (not w) => p -- Picked 2 p = if (not w) then True else False -- Contrapositive: (w => b) =equivalent-to=> ~b => ~w w = if (not b) then False else True b = False -- gives p is true and w is false

-- Second approach using pattern matching -- I think the rewriting goes from left to right but the logical inference goes in the opposite direction. w1, p1, b1::Bool p1 = (not w1) w1 = b1 -- Not consistent with statements, but I do not know how to write ~b1 => ~w1 in Haskell b1 = False -- Again gives p1 is true and w1 is false

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Alberto.

Patrick Browne

16 May 16 May

5:27 a.m.

Dan Mead

7:29 a.m.

maybe this will help? Haskell code in and of itself isn't special. proofs can happen with the type system, but typically you'd want to define a target language and do assertions about it, similar to how a compiler inspects it's input programs. Haskell is not homoiconic nor is it like coq or prolog. But it is really really good at defining an ast and doing operations on it. On Thu, May 16, 2013 at 5:27 AM, Patrick Browne wrote:

...

I do understand the difference between theorem provers and Haskell programs. Logic can be used to reason 'about' Haskell programs and logic can be used 'within' Haskell programs. I am trying to clarify the difference between 'about' and 'within' Is approach 1 concerned with |= (model based 'within'), whereas approach 2 is concerned with |- (proof based 'about')?

Thanks, Pat

On 15/05/13, *"Alberto G. Corona " * wrote:

Not exactly what you ask, but it is noteworthy that the mind has different logic processors. The fastest one work with IF THEN ELSE rules applied specifically to deals. This is why your example (and most examples of logic) involves a kind of deal expressed in the first person. This trigger a fast mental evaluation, while an equivalent but more general case is harder to process and need some paper work. (That special treatment of first person deals logic respond to the need to detect breaks of deals as fast as possible)

http://en.wikipedia.org/wiki/Wason_selection_task

That's why higher level languages have redundant logical structures and do not follow a general abstract and short mathematical notation. Therefore "higher level", in programming languages, does not mean higher mathematical abstraction, but to be closer to the way the mind works.

2013/5/15 Patrick Browne >

...
-- Hi -- I am trying to show that a set of propositions and a conclusion the form a valid argument. -- I used two approaches; 1) using if-then-else, 2) using pattern matching. -- The version using if-then-else seems to be consistent with my knowledge of Haskell and logic (either of which could be wrong). -- Can the second approach be improved to better reflect the propositions and conclusion? Maybe type level reasoning could be used? -- -- Valid argument? -- 1. I work hard or I play piano -- 2. If I work hard then I will get a bonus -- 3. But I did not get a bonus -- Therefore I played piano -- Variables: p = Piano, w = worked hard, b = got a bonus -- (w \/ p) /\ (w => b) /\ ¬(b) -- --------------------------- -- p

-- First approach using language control structure if-then-else w, p, b::Bool -- Two equivalences for (w \/ p) as an implication. -- 1. (w \/ p) =equivalent-to=> (not p) => w -- 2. (w \/ p) =equivalent-to=> (not w) => p -- Picked 2 p = if (not w) then True else False -- Contrapositive: (w => b) =equivalent-to=> ~b => ~w w = if (not b) then False else True b = False -- gives p is true and w is false

-- Second approach using pattern matching -- I think the rewriting goes from left to right but the logical inference goes in the opposite direction. w1, p1, b1::Bool p1 = (not w1) w1 = b1 -- Not consistent with statements, but I do not know how to write ~b1 => ~w1 in Haskell b1 = False -- Again gives p1 is true and w1 is false

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Alberto.

Tá an teachtaireacht seo scanta ó thaobh ábhar agus víreas ag Seirbhís Scanta Ríomhphost de chuid Seirbhísí Faisnéise, ITBÁC agus meastar í a bheith slán. http://www.dit.ie This message has been scanned for content and viruses by the DIT Information Services E-Mail Scanning Service, and is believed to be clean. http://www.dit.ie

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Keshav Kini

15 May 15 May

10:33 p.m.

Patrick Browne writes:

...

-- Hi

By the way, this is unrelated to your actual question, but if you haven't already, you might want to check out Bird-style Literate Haskell so you don't have to put "-- " in front of all of your non-Haskell text in a comment-heavy code-light file (or email message) :) http://www.haskell.org/haskellwiki/Literate_programming#Bird_Style -Keshav

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Alberto G. Corona
Dan Mead
Keshav Kini
MigMit
Patrick Browne
Tillmann Rendel