YES 0.863
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
↳ BR
mainModule Main
| ((inRange :: (Int,Int) -> Int -> Bool) :: (Int,Int) -> Int -> Bool) |
module Main where
Replaced joker patterns by fresh variables and removed binding patterns.
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
mainModule Main
| ((inRange :: (Int,Int) -> Int -> Bool) :: (Int,Int) -> Int -> Bool) |
module Main where
Cond Reductions:
The following Function with conditions
is transformed to
undefined0 | True | = undefined |
undefined1 | | = undefined0 False |
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
↳ HASKELL
↳ Narrow
mainModule Main
| (inRange :: (Int,Int) -> Int -> Bool) |
module Main where
Haskell To QDPs
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
↳ HASKELL
↳ Narrow
↳ AND
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
new_not(Succ(vx4000), Succ(vx31000)) → new_not(vx4000, vx31000)
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_not(Succ(vx4000), Succ(vx31000)) → new_not(vx4000, vx31000)
The graph contains the following edges 1 > 1, 2 > 2
↳ HASKELL
↳ BR
↳ HASKELL
↳ COR
↳ HASKELL
↳ Narrow
↳ AND
↳ QDP
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
new_asAs(Succ(vx300000), Succ(vx40000), vx6) → new_asAs(vx300000, vx40000, vx6)
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_asAs(Succ(vx300000), Succ(vx40000), vx6) → new_asAs(vx300000, vx40000, vx6)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3