/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.hs /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.hs
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


H-Termination with start terms of the given HASKELL could be proven:

(0) HASKELL
(1) BR [EQUIVALENT, 0 ms]
(2) HASKELL
(3) COR [EQUIVALENT, 0 ms]
(4) HASKELL
(5) Narrow [SOUND, 0 ms]
(6) QDP
(7) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(8) YES


----------------------------------------

(0)
Obligation:
mainModule Main 
module Main where {
import qualified Prelude;
data List a = Cons a (List a)  | Nil ;

scanl :: (b  ->  a  ->  b)  ->  b  ->  List a  ->  List b;
scanl f q xs = Cons q (scanl0 f q xs);

scanl0 f q Nil = Nil;
scanl0 f q (Cons x xs) = scanl f (f q x) xs;

scanl1 :: (a  ->  a  ->  a)  ->  List a  ->  List a;
scanl1 vv Nil = Nil;
scanl1 f (Cons x xs) = scanl f x xs;

}

----------------------------------------

(1) BR (EQUIVALENT)
Replaced joker patterns by fresh variables and removed binding patterns.
----------------------------------------

(2)
Obligation:
mainModule Main 
module Main where {
import qualified Prelude;
data List a = Cons a (List a)  | Nil ;

scanl :: (b  ->  a  ->  b)  ->  b  ->  List a  ->  List b;
scanl f q xs = Cons q (scanl0 f q xs);

scanl0 f q Nil = Nil;
scanl0 f q (Cons x xs) = scanl f (f q x) xs;

scanl1 :: (a  ->  a  ->  a)  ->  List a  ->  List a;
scanl1 vv Nil = Nil;
scanl1 f (Cons x xs) = scanl f x xs;

}

----------------------------------------

(3) COR (EQUIVALENT)
Cond Reductions:
The following Function with conditions
"undefined |Falseundefined;
"
is transformed to
"undefined  = undefined1;
"
"undefined0 True = undefined;
"
"undefined1  = undefined0 False;
"

----------------------------------------

(4)
Obligation:
mainModule Main 
module Main where {
import qualified Prelude;
data List a = Cons a (List a)  | Nil ;

scanl :: (a  ->  b  ->  a)  ->  a  ->  List b  ->  List a;
scanl f q xs = Cons q (scanl0 f q xs);

scanl0 f q Nil = Nil;
scanl0 f q (Cons x xs) = scanl f (f q x) xs;

scanl1 :: (a  ->  a  ->  a)  ->  List a  ->  List a;
scanl1 vv Nil = Nil;
scanl1 f (Cons x xs) = scanl f x xs;

}

----------------------------------------

(5) Narrow (SOUND)
Haskell To QDPs

digraph dp_graph {
node [outthreshold=100, inthreshold=100];1[label="scanl1",fontsize=16,color="grey",shape="box"];1 -> 3[label="",style="dashed", color="grey", weight=3];
3[label="scanl1 vy3",fontsize=16,color="grey",shape="box"];3 -> 4[label="",style="dashed", color="grey", weight=3];
4[label="scanl1 vy3 vy4",fontsize=16,color="burlywood",shape="triangle"];19[label="vy4/Cons vy40 vy41",fontsize=10,color="white",style="solid",shape="box"];4 -> 19[label="",style="solid", color="burlywood", weight=9];
19 -> 5[label="",style="solid", color="burlywood", weight=3];
20[label="vy4/Nil",fontsize=10,color="white",style="solid",shape="box"];4 -> 20[label="",style="solid", color="burlywood", weight=9];
20 -> 6[label="",style="solid", color="burlywood", weight=3];
5[label="scanl1 vy3 (Cons vy40 vy41)",fontsize=16,color="black",shape="box"];5 -> 7[label="",style="solid", color="black", weight=3];
6[label="scanl1 vy3 Nil",fontsize=16,color="black",shape="box"];6 -> 8[label="",style="solid", color="black", weight=3];
7[label="scanl vy3 vy40 vy41",fontsize=16,color="black",shape="triangle"];7 -> 9[label="",style="solid", color="black", weight=3];
8[label="Nil",fontsize=16,color="green",shape="box"];9[label="Cons vy40 (scanl0 vy3 vy40 vy41)",fontsize=16,color="green",shape="box"];9 -> 10[label="",style="dashed", color="green", weight=3];
10[label="scanl0 vy3 vy40 vy41",fontsize=16,color="burlywood",shape="box"];21[label="vy41/Cons vy410 vy411",fontsize=10,color="white",style="solid",shape="box"];10 -> 21[label="",style="solid", color="burlywood", weight=9];
21 -> 11[label="",style="solid", color="burlywood", weight=3];
22[label="vy41/Nil",fontsize=10,color="white",style="solid",shape="box"];10 -> 22[label="",style="solid", color="burlywood", weight=9];
22 -> 12[label="",style="solid", color="burlywood", weight=3];
11[label="scanl0 vy3 vy40 (Cons vy410 vy411)",fontsize=16,color="black",shape="box"];11 -> 13[label="",style="solid", color="black", weight=3];
12[label="scanl0 vy3 vy40 Nil",fontsize=16,color="black",shape="box"];12 -> 14[label="",style="solid", color="black", weight=3];
13 -> 7[label="",style="dashed", color="red", weight=0];
13[label="scanl vy3 (vy3 vy40 vy410) vy411",fontsize=16,color="magenta"];13 -> 15[label="",style="dashed", color="magenta", weight=3];
13 -> 16[label="",style="dashed", color="magenta", weight=3];
14[label="Nil",fontsize=16,color="green",shape="box"];15[label="vy3 vy40 vy410",fontsize=16,color="green",shape="box"];15 -> 17[label="",style="dashed", color="green", weight=3];
15 -> 18[label="",style="dashed", color="green", weight=3];
16[label="vy411",fontsize=16,color="green",shape="box"];17[label="vy40",fontsize=16,color="green",shape="box"];18[label="vy410",fontsize=16,color="green",shape="box"];}

----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   new_scanl(vy3, Cons(vy410, vy411), h) -> new_scanl(vy3, vy411, h)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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_scanl(vy3, Cons(vy410, vy411), h) -> new_scanl(vy3, vy411, h)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


----------------------------------------

(8)
YES