YES 6.907 H-Termination proof of /home/matraf/haskell/eval_FullyBlown_Fast/empty.hs
H-Termination of the given Haskell-Program with start terms could successfully be proven:



HASKELL
  ↳ IFR

mainModule Main
  ((realToFrac :: Int  ->  Float) :: Int  ->  Float)

module Main where
  import qualified Prelude



If Reductions:
The following If expression
if primGEqNatS x y then Succ (primDivNatS (primMinusNatS x y) (Succ y)) else Zero

is transformed to
primDivNatS0 x y True = Succ (primDivNatS (primMinusNatS x y) (Succ y))
primDivNatS0 x y False = Zero

The following If expression
if primGEqNatS x y then primModNatS (primMinusNatS x y) (Succ y) else Succ x

is transformed to
primModNatS0 x y True = primModNatS (primMinusNatS x y) (Succ y)
primModNatS0 x y False = Succ x



↳ HASKELL
  ↳ IFR
HASKELL
      ↳ BR

mainModule Main
  ((realToFrac :: Int  ->  Float) :: Int  ->  Float)

module Main where
  import qualified Prelude



Replaced joker patterns by fresh variables and removed binding patterns.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
HASKELL
          ↳ COR

mainModule Main
  ((realToFrac :: Int  ->  Float) :: Int  ->  Float)

module Main where
  import qualified Prelude



Cond Reductions:
The following Function with conditions
gcd' x 0 = x
gcd' x y = gcd' y (x `rem` y)

is transformed to
gcd' x xz = gcd'2 x xz
gcd' x y = gcd'0 x y

gcd'0 x y = gcd' y (x `rem` y)

gcd'1 True x xz = x
gcd'1 yu yv yw = gcd'0 yv yw

gcd'2 x xz = gcd'1 (xz == 0) x xz
gcd'2 yx yy = gcd'0 yx yy

The following Function with conditions
gcd 0 0 = error []
gcd x y = 
gcd' (abs x) (abs y)
where 
gcd' x 0 = x
gcd' x y = gcd' y (x `rem` y)

is transformed to
gcd yz zu = gcd3 yz zu
gcd x y = gcd0 x y

gcd0 x y = 
gcd' (abs x) (abs y)
where 
gcd' x xz = gcd'2 x xz
gcd' x y = gcd'0 x y
gcd'0 x y = gcd' y (x `rem` y)
gcd'1 True x xz = x
gcd'1 yu yv yw = gcd'0 yv yw
gcd'2 x xz = gcd'1 (xz == 0) x xz
gcd'2 yx yy = gcd'0 yx yy

gcd1 True yz zu = error []
gcd1 zv zw zx = gcd0 zw zx

gcd2 True yz zu = gcd1 (zu == 0) yz zu
gcd2 zy zz vuu = gcd0 zz vuu

gcd3 yz zu = gcd2 (yz == 0) yz zu
gcd3 vuv vuw = gcd0 vuv vuw

The following Function with conditions
absReal x
 | x >= 0
 = x
 | otherwise
 = `negate` x

is transformed to
absReal x = absReal2 x

absReal1 x True = x
absReal1 x False = absReal0 x otherwise

absReal0 x True = `negate` x

absReal2 x = absReal1 x (x >= 0)

The following Function with conditions
undefined 
 | False
 = undefined

is transformed to
undefined  = undefined1

undefined0 True = undefined

undefined1  = undefined0 False

The following Function with conditions
reduce x y
 | y == 0
 = error []
 | otherwise
 = x `quot` d :% (y `quot` d)
where 
d  = gcd x y

is transformed to
reduce x y = reduce2 x y

reduce2 x y = 
reduce1 x y (y == 0)
where 
d  = gcd x y
reduce0 x y True = x `quot` d :% (y `quot` d)
reduce1 x y True = error []
reduce1 x y False = reduce0 x y otherwise

The following Function with conditions
signumReal x
 | x == 0
 = 0
 | x > 0
 = 1
 | otherwise
 = -1

is transformed to
signumReal x = signumReal3 x

signumReal0 x True = -1

signumReal2 x True = 0
signumReal2 x False = signumReal1 x (x > 0)

signumReal1 x True = 1
signumReal1 x False = signumReal0 x otherwise

signumReal3 x = signumReal2 x (x == 0)



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
HASKELL
              ↳ LetRed

mainModule Main
  ((realToFrac :: Int  ->  Float) :: Int  ->  Float)

module Main where
  import qualified Prelude



Let/Where Reductions:
The bindings of the following Let/Where expression
reduce1 x y (y == 0)
where 
d  = gcd x y
reduce0 x y True = x `quot` d :% (y `quot` d)
reduce1 x y True = error []
reduce1 x y False = reduce0 x y otherwise

are unpacked to the following functions on top level
reduce2Reduce0 vux vuy x y True = x `quot` reduce2D vux vuy :% (y `quot` reduce2D vux vuy)

reduce2Reduce1 vux vuy x y True = error []
reduce2Reduce1 vux vuy x y False = reduce2Reduce0 vux vuy x y otherwise

reduce2D vux vuy = gcd vux vuy

The bindings of the following Let/Where expression
gcd' (abs x) (abs y)
where 
gcd' x xz = gcd'2 x xz
gcd' x y = gcd'0 x y
gcd'0 x y = gcd' y (x `rem` y)
gcd'1 True x xz = x
gcd'1 yu yv yw = gcd'0 yv yw
gcd'2 x xz = gcd'1 (xz == 0) x xz
gcd'2 yx yy = gcd'0 yx yy

are unpacked to the following functions on top level
gcd0Gcd'2 x xz = gcd0Gcd'1 (xz == 0) x xz
gcd0Gcd'2 yx yy = gcd0Gcd'0 yx yy

gcd0Gcd'1 True x xz = x
gcd0Gcd'1 yu yv yw = gcd0Gcd'0 yv yw

gcd0Gcd' x xz = gcd0Gcd'2 x xz
gcd0Gcd' x y = gcd0Gcd'0 x y

gcd0Gcd'0 x y = gcd0Gcd' y (x `rem` y)



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
HASKELL
                  ↳ NumRed

mainModule Main
  ((realToFrac :: Int  ->  Float) :: Int  ->  Float)

module Main where
  import qualified Prelude



Num Reduction: All numbers are transformed to thier corresponding representation with Pos, Neg, Succ and Zero.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
HASKELL
                      ↳ Narrow

mainModule Main
  (realToFrac :: Int  ->  Float)

module Main where
  import qualified Prelude



Haskell To QDPs


↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
QDP
                            ↳ QDPSizeChangeProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

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

new_primMulNat(Succ(vuz3000)) → new_primMulNat(vuz3000)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
QDP
                            ↳ DependencyGraphProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

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

new_primDivNatS(Succ(Zero)) → new_primDivNatS(new_primMinusNatS0)
new_primDivNatS(Succ(Succ(vuz770000))) → new_primDivNatS(new_primMinusNatS(vuz770000))

The TRS R consists of the following rules:

new_primMinusNatS0Zero
new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ UsableRulesProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

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

new_primDivNatS(Succ(Succ(vuz770000))) → new_primDivNatS(new_primMinusNatS(vuz770000))

The TRS R consists of the following rules:

new_primMinusNatS0Zero
new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

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

new_primDivNatS(Succ(Succ(vuz770000))) → new_primDivNatS(new_primMinusNatS(vuz770000))

The TRS R consists of the following rules:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

new_primMinusNatS0



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP
                                        ↳ RuleRemovalProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

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

new_primDivNatS(Succ(Succ(vuz770000))) → new_primDivNatS(new_primMinusNatS(vuz770000))

The TRS R consists of the following rules:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

new_primDivNatS(Succ(Succ(vuz770000))) → new_primDivNatS(new_primMinusNatS(vuz770000))

Strictly oriented rules of the TRS R:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

Used ordering: POLO with Polynomial interpretation [25]:

POL(Succ(x1)) = 1 + 2·x1   
POL(new_primDivNatS(x1)) = x1   
POL(new_primMinusNatS(x1)) = 2 + 2·x1   



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
                                      ↳ QDP
                                        ↳ RuleRemovalProof
QDP
                                            ↳ PisEmptyProof
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

new_primMinusNatS(x0)

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

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
QDP
                            ↳ DependencyGraphProof
                          ↳ QDP
                          ↳ QDP

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

new_quot(vuz72, Succ(Zero), vuz78) → new_quot(vuz72, new_primMinusNatS0, new_primMinusNatS0)
new_quot(vuz72, Succ(Succ(vuz7900)), vuz78) → new_quot(vuz72, new_primMinusNatS(vuz7900), new_primMinusNatS(vuz7900))

The TRS R consists of the following rules:

new_primMinusNatS0Zero
new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ UsableRulesProof
                          ↳ QDP
                          ↳ QDP

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

new_quot(vuz72, Succ(Succ(vuz7900)), vuz78) → new_quot(vuz72, new_primMinusNatS(vuz7900), new_primMinusNatS(vuz7900))

The TRS R consists of the following rules:

new_primMinusNatS0Zero
new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof
                          ↳ QDP
                          ↳ QDP

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

new_quot(vuz72, Succ(Succ(vuz7900)), vuz78) → new_quot(vuz72, new_primMinusNatS(vuz7900), new_primMinusNatS(vuz7900))

The TRS R consists of the following rules:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS0
new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

new_primMinusNatS0



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP
                                        ↳ RuleRemovalProof
                          ↳ QDP
                          ↳ QDP

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

new_quot(vuz72, Succ(Succ(vuz7900)), vuz78) → new_quot(vuz72, new_primMinusNatS(vuz7900), new_primMinusNatS(vuz7900))

The TRS R consists of the following rules:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS(x0)

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

new_quot(vuz72, Succ(Succ(vuz7900)), vuz78) → new_quot(vuz72, new_primMinusNatS(vuz7900), new_primMinusNatS(vuz7900))


Used ordering: POLO with Polynomial interpretation [25]:

POL(Succ(x1)) = 1 + 2·x1   
POL(new_primMinusNatS(x1)) = 1 + 2·x1   
POL(new_quot(x1, x2, x3)) = x1 + x2 + x3   



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
                                      ↳ QDP
                                        ↳ RuleRemovalProof
QDP
                                            ↳ PisEmptyProof
                          ↳ QDP
                          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

new_primMinusNatS(vuz770000) → Succ(vuz770000)

The set Q consists of the following terms:

new_primMinusNatS(x0)

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

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
QDP
                            ↳ QDPSizeChangeProof
                          ↳ QDP

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

new_quot0(Succ(vuz4400)) → new_quot0(vuz4400)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ LetRed
                ↳ HASKELL
                  ↳ NumRed
                    ↳ HASKELL
                      ↳ Narrow
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
                          ↳ QDP
QDP
                            ↳ QDPSizeChangeProof

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

new_quot1(vuz74, Succ(vuz750)) → new_quot1(vuz74, vuz750)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: