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


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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 4 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) ATransformationProof [EQUIVALENT, 1 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) ATransformationProof [EQUIVALENT, 1 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES
    (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   app(intlist, nil) -> nil
   app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y))
   app(app(int, 0), 0) -> app(app(cons, 0), nil)
   app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   app(app(int, app(s, x)), 0) -> nil
   app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)


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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(intlist, app(app(cons, x), y)) -> APP(app(cons, app(s, x)), app(intlist, y))
   APP(intlist, app(app(cons, x), y)) -> APP(cons, app(s, x))
   APP(intlist, app(app(cons, x), y)) -> APP(s, x)
   APP(intlist, app(app(cons, x), y)) -> APP(intlist, y)
   APP(app(int, 0), 0) -> APP(app(cons, 0), nil)
   APP(app(int, 0), 0) -> APP(cons, 0)
   APP(app(int, 0), app(s, y)) -> APP(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   APP(app(int, 0), app(s, y)) -> APP(cons, 0)
   APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y))
   APP(app(int, 0), app(s, y)) -> APP(int, app(s, 0))
   APP(app(int, 0), app(s, y)) -> APP(s, 0)
   APP(app(int, app(s, x)), app(s, y)) -> APP(intlist, app(app(int, x), y))
   APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y)
   APP(app(int, app(s, x)), app(s, y)) -> APP(int, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x))
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs))
   APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f)

The TRS R consists of the following rules:

   app(intlist, nil) -> nil
   app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y))
   app(app(int, 0), 0) -> app(app(cons, 0), nil)
   app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   app(app(int, app(s, x)), 0) -> nil
   app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 21 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(intlist, app(app(cons, x), y)) -> APP(intlist, y)

The TRS R consists of the following rules:

   app(intlist, nil) -> nil
   app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y))
   app(app(int, 0), 0) -> app(app(cons, 0), nil)
   app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   app(app(int, app(s, x)), 0) -> nil
   app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(6) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(intlist, app(app(cons, x), y)) -> APP(intlist, y)

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

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(8) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   intlist1(cons(x, y)) -> intlist1(y)

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

   intlist(nil)
   intlist(cons(x0, x1))
   int(0, 0)
   int(0, s(x0))
   int(s(x0), 0)
   int(s(x0), s(x1))
   map(x0, nil)
   map(x0, cons(x1, x2))
   filter(x0, nil)
   filter(x0, cons(x1, x2))
   filter2(true, x0, x1, x2)
   filter2(false, x0, x1, x2)

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   intlist(nil)
   intlist(cons(x0, x1))
   int(0, 0)
   int(0, s(x0))
   int(s(x0), 0)
   int(s(x0), s(x1))
   map(x0, nil)
   map(x0, cons(x1, x2))
   filter(x0, nil)
   filter(x0, cons(x1, x2))
   filter2(true, x0, x1, x2)
   filter2(false, x0, x1, x2)


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   intlist1(cons(x, y)) -> intlist1(y)

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

(12) 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:
*intlist1(cons(x, y)) -> intlist1(y)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y)
   APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y))

The TRS R consists of the following rules:

   app(intlist, nil) -> nil
   app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y))
   app(app(int, 0), 0) -> app(app(cons, 0), nil)
   app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   app(app(int, app(s, x)), 0) -> nil
   app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y)
   APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y))

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

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(17) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   int1(s(x), s(y)) -> int1(x, y)
   int1(0, s(y)) -> int1(s(0), s(y))

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

   intlist(nil)
   intlist(cons(x0, x1))
   int(0, 0)
   int(0, s(x0))
   int(s(x0), 0)
   int(s(x0), s(x1))
   map(x0, nil)
   map(x0, cons(x1, x2))
   filter(x0, nil)
   filter(x0, cons(x1, x2))
   filter2(true, x0, x1, x2)
   filter2(false, x0, x1, x2)

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

(19) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   intlist(nil)
   intlist(cons(x0, x1))
   int(0, 0)
   int(0, s(x0))
   int(s(x0), 0)
   int(s(x0), s(x1))
   map(x0, nil)
   map(x0, cons(x1, x2))
   filter(x0, nil)
   filter(x0, cons(x1, x2))
   filter2(true, x0, x1, x2)
   filter2(false, x0, x1, x2)


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   int1(s(x), s(y)) -> int1(x, y)
   int1(0, s(y)) -> int1(s(0), s(y))

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

(21) 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:
*int1(s(x), s(y)) -> int1(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*int1(0, s(y)) -> int1(s(0), s(y))
The graph contains the following edges 2 >= 2


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

(22)
YES

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

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

The TRS R consists of the following rules:

   app(intlist, nil) -> nil
   app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y))
   app(app(int, 0), 0) -> app(app(cons, 0), nil)
   app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y)))
   app(app(int, app(s, x)), 0) -> nil
   app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(24) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

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

   app(intlist, nil)
   app(intlist, app(app(cons, x0), x1))
   app(app(int, 0), 0)
   app(app(int, 0), app(s, x0))
   app(app(int, app(s, x0)), 0)
   app(app(int, app(s, x0)), app(s, x1))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

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

(26) 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:
*APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2


*APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2


*APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2


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(27)
YES