MAYBE 1.395 H-Termination proof of /home/matraf/haskell/eval_FullyBlown_Fast/empty.hs
H-Termination of the given Haskell-Program with start terms could not be shown:



HASKELL
  ↳ IFR

mainModule Main
  ((until :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a) :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a)

module Main where
  import qualified Prelude



If Reductions:
The following If expression
if p x then x else until p f (f x)

is transformed to
until0 x p f True = x
until0 x p f False = until p f (f x)



↳ HASKELL
  ↳ IFR
HASKELL
      ↳ BR

mainModule Main
  ((until :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a) :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a)

module Main where
  import qualified Prelude



Replaced joker patterns by fresh variables and removed binding patterns.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
HASKELL
          ↳ COR

mainModule Main
  ((until :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a) :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a)

module Main where
  import qualified Prelude



Cond Reductions:
The following Function with conditions
undefined 
 | False
 = undefined

is transformed to
undefined  = undefined1

undefined0 True = undefined

undefined1  = undefined0 False



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
HASKELL
              ↳ Narrow
              ↳ Narrow

mainModule Main
  (until :: (a  ->  Bool ->  (a  ->  a ->  a  ->  a)

module Main where
  import qualified Prelude



Haskell To QDPs


↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ Narrow
QDP
                  ↳ NonTerminationProof
              ↳ Narrow

Q DP problem:
The TRS P consists of the following rules:

new_until(vx3, vx4, ba) → new_until0(vx3, vx4, ba)
new_until0(vx3, vx4, ba) → new_until(vx3, vx4, ba)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

new_until(vx3, vx4, ba) → new_until0(vx3, vx4, ba)
new_until0(vx3, vx4, ba) → new_until(vx3, vx4, ba)

The TRS R consists of the following rules:none


s = new_until0(vx3', vx4', ba') evaluates to t =new_until0(vx3', vx4', ba')

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

new_until0(vx3', vx4', ba')new_until(vx3', vx4', ba')
with rule new_until0(vx3'', vx4'', ba'') → new_until(vx3'', vx4'', ba'') at position [] and matcher [ba'' / ba', vx4'' / vx4', vx3'' / vx3']

new_until(vx3', vx4', ba')new_until0(vx3', vx4', ba')
with rule new_until(vx3, vx4, ba) → new_until0(vx3, vx4, ba)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




Haskell To QDPs


↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ Narrow
              ↳ Narrow
QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.