YES 1.6260000000000001
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
| ((quot :: Int -> Int -> Int) :: Int -> Int -> Int) |
module Main where
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 |
↳ HASKELL
↳ IFR
↳ HASKELL
↳ BR
mainModule Main
| ((quot :: Int -> Int -> Int) :: Int -> Int -> Int) |
module Main where
Replaced joker patterns by fresh variables and removed binding patterns.
↳ HASKELL
↳ IFR
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
mainModule Main
| ((quot :: Int -> Int -> Int) :: Int -> Int -> Int) |
module Main where
Cond Reductions:
The following Function with conditions
is transformed to
undefined0 | True | = undefined |
undefined1 | | = undefined0 False |
↳ HASKELL
↳ IFR
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
↳ HASKELL
↳ Narrow
mainModule Main
| (quot :: Int -> Int -> Int) |
module Main where
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:
- new_primMinusNatS(Succ(vz210), Succ(vz220)) → new_primMinusNatS(vz210, vz220)
The graph contains the following edges 1 > 1, 2 > 2
↳ HASKELL
↳ IFR
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
↳ HASKELL
↳ Narrow
↳ AND
↳ QDP
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
new_primDivNatS0(vz21, vz22, Zero, Zero) → new_primDivNatS00(vz21, vz22)
new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primDivNatS0(vz3000, vz4000, vz3000, vz4000)
new_primDivNatS0(vz21, vz22, Succ(vz230), Zero) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS(Succ(Zero), Zero) → new_primDivNatS(new_primMinusNatS2, Zero)
new_primDivNatS(Succ(Succ(vz3000)), Zero) → new_primDivNatS(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_primMinusNatS2 → Zero
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_primDivNatS(Succ(Succ(vz3000)), Zero) → new_primDivNatS(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_primMinusNatS2 → Zero
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_primDivNatS(Succ(Succ(vz3000)), Zero) → new_primDivNatS(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_primDivNatS(Succ(Succ(vz3000)), Zero) → new_primDivNatS(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_primDivNatS(Succ(Succ(vz3000)), Zero) → new_primDivNatS(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_primDivNatS(x1, x2)) = x1 + x2
POL(new_primMinusNatS1(x1)) = 2 + 2·x1
↳ 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_primDivNatS0(vz21, vz22, Zero, Zero) → new_primDivNatS00(vz21, vz22)
new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
new_primDivNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primDivNatS0(vz3000, vz4000, vz3000, vz4000)
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS0(vz21, vz22, Succ(vz230), Zero) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(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_primMinusNatS2 → Zero
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_primDivNatS0(vz21, vz22, Zero, Zero) → new_primDivNatS00(vz21, vz22)
new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
new_primDivNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primDivNatS0(vz3000, vz4000, vz3000, vz4000)
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS0(vz21, vz22, Succ(vz230), Zero) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(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_primDivNatS0(vz21, vz22, Zero, Zero) → new_primDivNatS00(vz21, vz22)
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primDivNatS0(vz3000, vz4000, vz3000, vz4000)
new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
new_primDivNatS0(vz21, vz22, Succ(vz230), Zero) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(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.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
new_primDivNatS0(vz21, vz22, Zero, Zero) → new_primDivNatS00(vz21, vz22)
new_primDivNatS(Succ(Succ(vz3000)), Succ(vz4000)) → new_primDivNatS0(vz3000, vz4000, vz3000, vz4000)
new_primDivNatS0(vz21, vz22, Succ(vz230), Zero) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
The remaining pairs can at least be oriented weakly.
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(vz22))
new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
Used ordering: Polynomial interpretation [25]:
POL(Succ(x1)) = 1 + x1
POL(Zero) = 0
POL(new_primDivNatS(x1, x2)) = x1
POL(new_primDivNatS0(x1, x2, x3, x4)) = 1 + x1
POL(new_primDivNatS00(x1, x2)) = x1
POL(new_primMinusNatS0(x1, x2)) = 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_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
new_primDivNatS00(vz21, vz22) → new_primDivNatS(new_primMinusNatS0(vz21, vz22), Succ(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_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(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_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(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_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(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:
- new_primDivNatS0(vz21, vz22, Succ(vz230), Succ(vz240)) → new_primDivNatS0(vz21, vz22, vz230, vz240)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3, 4 > 4