/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 1 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) ATransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) ATransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES


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

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

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

The set Q consists of the following terms:

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))


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

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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
   APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
   APP(app(plus, app(s, x)), y) -> APP(plus, x)
   APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
   APP(app(times, app(s, x)), y) -> APP(plus, app(app(times, x), y))
   APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
   APP(app(times, app(s, x)), y) -> APP(times, x)
   APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
   APP(inc, xs) -> APP(map, app(plus, app(s, 0)))
   APP(inc, xs) -> APP(plus, app(s, 0))
   APP(inc, xs) -> APP(s, 0)
   APP(double, xs) -> APP(app(map, app(times, app(s, app(s, 0)))), xs)
   APP(double, xs) -> APP(map, app(times, app(s, app(s, 0))))
   APP(double, xs) -> APP(times, app(s, app(s, 0)))
   APP(double, xs) -> APP(s, app(s, 0))
   APP(double, xs) -> APP(s, 0)
   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)

The TRS R consists of the following rules:

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

The set Q consists of the following terms:

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 14 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)

The TRS R consists of the following rules:

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

The set Q consists of the following terms:

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

(8) 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.
----------------------------------------

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

   APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)

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

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

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

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

   plus1(s(x), y) -> plus1(x, y)

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

   plus(0, x0)
   plus(s(x0), x1)
   times(0, x0)
   times(s(x0), x1)
   inc(x0)
   double(x0)
   map(x0, nil)
   map(x0, cons(x1, x2))

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

(12) 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].

   plus(0, x0)
   plus(s(x0), x1)
   times(0, x0)
   times(s(x0), x1)
   inc(x0)
   double(x0)
   map(x0, nil)
   map(x0, cons(x1, x2))


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   plus1(s(x), y) -> plus1(x, y)

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

(14) 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:
*plus1(s(x), y) -> plus1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(15)
YES

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

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

   APP(app(times, app(s, x)), y) -> APP(app(times, x), y)

The TRS R consists of the following rules:

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

The set Q consists of the following terms:

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

(17) 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.
----------------------------------------

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

   APP(app(times, app(s, x)), y) -> APP(app(times, x), y)

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

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

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

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

   times1(s(x), y) -> times1(x, y)

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

   plus(0, x0)
   plus(s(x0), x1)
   times(0, x0)
   times(s(x0), x1)
   inc(x0)
   double(x0)
   map(x0, nil)
   map(x0, cons(x1, x2))

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

(21) 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].

   plus(0, x0)
   plus(s(x0), x1)
   times(0, x0)
   times(s(x0), x1)
   inc(x0)
   double(x0)
   map(x0, nil)
   map(x0, cons(x1, x2))


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   times1(s(x), y) -> times1(x, y)

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

(23) 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:
*times1(s(x), y) -> times1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(24)
YES

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

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

   APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(double, xs) -> APP(app(map, app(times, app(s, app(s, 0)))), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)

The TRS R consists of the following rules:

   app(app(plus, 0), y) -> y
   app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
   app(app(times, 0), y) -> 0
   app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
   app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
   app(double, xs) -> app(app(map, app(times, app(s, app(s, 0)))), xs)
   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))

The set Q consists of the following terms:

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

(26) 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.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(double, xs) -> APP(app(map, app(times, app(s, app(s, 0)))), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)

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

   app(app(plus, 0), x0)
   app(app(plus, app(s, x0)), x1)
   app(app(times, 0), x0)
   app(app(times, app(s, x0)), x1)
   app(inc, x0)
   app(double, x0)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))

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

(28) 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(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(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
The graph contains the following edges 2 >= 2


*APP(double, xs) -> APP(app(map, app(times, app(s, app(s, 0)))), xs)
The graph contains the following edges 2 >= 2


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

(29)
YES