YES 1.586 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
  ((rem :: Int  ->  Int  ->  Int) :: Int  ->  Int  ->  Int)

module Main where
  import qualified Prelude



If Reductions:
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
  ((rem :: Int  ->  Int  ->  Int) :: Int  ->  Int  ->  Int)

module Main where
  import qualified Prelude



Replaced joker patterns by fresh variables and removed binding patterns.

↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
HASKELL
          ↳ COR

mainModule Main
  ((rem :: Int  ->  Int  ->  Int) :: Int  ->  Int  ->  Int)

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

mainModule Main
  (rem :: Int  ->  Int  ->  Int)

module Main where
  import qualified Prelude



Haskell To QDPs


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

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

new_primMinusNatS(Succ(vz210), Succ(vz220)) → new_primMinusNatS(vz210, vz220)

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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
QDP
                    ↳ DependencyGraphProof

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

new_primModNatS00(vz21, vz22) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primModNatS0(vz3000, vz4000, vz3000, vz4000)
new_primModNatS(Succ(Zero), Zero) → new_primModNatS(new_primMinusNatS2, Zero)
new_primModNatS0(vz21, vz22, Succ(vz230), Zero) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Zero) → new_primModNatS(new_primMinusNatS1(vz3000), Zero)
new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)
new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS1(vz3000) → Succ(vz3000)
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)
new_primMinusNatS2Zero

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS1(x0)
new_primMinusNatS0(Zero, Zero)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

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

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

new_primModNatS(Succ(Succ(vz3000)), Zero) → new_primModNatS(new_primMinusNatS1(vz3000), Zero)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS1(vz3000) → Succ(vz3000)
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)
new_primMinusNatS2Zero

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS1(x0)
new_primMinusNatS0(Zero, Zero)

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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                          ↳ UsableRulesProof
QDP
                              ↳ QReductionProof
                        ↳ QDP

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

new_primModNatS(Succ(Succ(vz3000)), Zero) → new_primModNatS(new_primMinusNatS1(vz3000), Zero)

The TRS R consists of the following rules:

new_primMinusNatS1(vz3000) → Succ(vz3000)

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS1(x0)
new_primMinusNatS0(Zero, Zero)

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(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)



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

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

new_primModNatS(Succ(Succ(vz3000)), Zero) → new_primModNatS(new_primMinusNatS1(vz3000), Zero)

The TRS R consists of the following rules:

new_primMinusNatS1(vz3000) → Succ(vz3000)

The set Q consists of the following terms:

new_primMinusNatS1(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_primModNatS(Succ(Succ(vz3000)), Zero) → new_primModNatS(new_primMinusNatS1(vz3000), Zero)

Strictly oriented rules of the TRS R:

new_primMinusNatS1(vz3000) → Succ(vz3000)

Used ordering: POLO with Polynomial interpretation [25]:

POL(Succ(x1)) = 1 + 2·x1   
POL(Zero) = 0   
POL(new_primMinusNatS1(x1)) = 2 + 2·x1   
POL(new_primModNatS(x1, x2)) = x1 + x2   



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

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

new_primMinusNatS1(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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
QDP
                          ↳ UsableRulesProof

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

new_primModNatS00(vz21, vz22) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primModNatS0(vz3000, vz4000, vz3000, vz4000)
new_primModNatS0(vz21, vz22, Succ(vz230), Zero) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)
new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS1(vz3000) → Succ(vz3000)
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)
new_primMinusNatS2Zero

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS1(x0)
new_primMinusNatS0(Zero, Zero)

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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
QDP
                              ↳ QReductionProof

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

new_primModNatS00(vz21, vz22) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primModNatS0(vz3000, vz4000, vz3000, vz4000)
new_primModNatS0(vz21, vz22, Succ(vz230), Zero) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)
new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS2
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS1(x0)
new_primMinusNatS0(Zero, Zero)

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_primMinusNatS2
new_primMinusNatS1(x0)



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
QDP
                                  ↳ QDPOrderProof

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

new_primModNatS00(vz21, vz22) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primModNatS0(vz3000, vz4000, vz3000, vz4000)
new_primModNatS0(vz21, vz22, Succ(vz230), Zero) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)
new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


new_primModNatS00(vz21, vz22) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primModNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primModNatS0(vz3000, vz4000, vz3000, vz4000)
new_primModNatS0(vz21, vz22, Succ(vz230), Zero) → new_primModNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
The remaining pairs can at least be oriented weakly.

new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)
new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)
Used ordering: Polynomial interpretation [25]:

POL(Succ(x1)) = 1 + x1   
POL(Zero) = 0   
POL(new_primMinusNatS0(x1, x2)) = x1   
POL(new_primModNatS(x1, x2)) = x1   
POL(new_primModNatS0(x1, x2, x3, x4)) = 1 + x1   
POL(new_primModNatS00(x1, x2)) = 1 + x1   

The following usable rules [17] were oriented:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ QDPOrderProof
QDP
                                      ↳ DependencyGraphProof

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

new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)
new_primModNatS0(vz21, vz22, Zero, Zero) → new_primModNatS00(vz21, vz22)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)

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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ QDPOrderProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                                          ↳ UsableRulesProof

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

new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)

The TRS R consists of the following rules:

new_primMinusNatS0(Zero, Succ(vz220)) → Zero
new_primMinusNatS0(Zero, Zero) → Zero
new_primMinusNatS0(Succ(vz210), Succ(vz220)) → new_primMinusNatS0(vz210, vz220)
new_primMinusNatS0(Succ(vz210), Zero) → Succ(vz210)

The set Q consists of the following terms:

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)

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
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ QDPOrderProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
QDP
                                              ↳ QReductionProof

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

new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)

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

new_primMinusNatS0(Succ(x0), Zero)
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)

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(Succ(x0), Zero)
new_primMinusNatS0(Succ(x0), Succ(x1))
new_primMinusNatS0(Zero, Succ(x0))
new_primMinusNatS0(Zero, Zero)



↳ HASKELL
  ↳ IFR
    ↳ HASKELL
      ↳ BR
        ↳ HASKELL
          ↳ COR
            ↳ HASKELL
              ↳ Narrow
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ QDPOrderProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
QDP
                                                  ↳ QDPSizeChangeProof

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

new_primModNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primModNatS0(vz21, vz22, vz230, vz240)

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: