/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 3 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) ATransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES


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

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

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   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)

Q is empty.

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

(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(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))
   APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
   APP(app(., app(app(., x), y)), z) -> APP(., y)
   APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
   APP(i, app(app(., x), y)) -> APP(., app(i, y))
   APP(i, app(app(., x), y)) -> APP(i, y)
   APP(i, app(app(., x), y)) -> APP(i, 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(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   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)

Q is empty.
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 12 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
   APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   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)

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
   APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))

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

(8) UsableRulesReductionPairsProof (EQUIVALENT)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.

The pairs P are: 

   .1(.(x, y), z) -> .1(y, z)
   .1(.(x, y), z) -> .1(x, .(y, z))

and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:

   .(1, x) -> x
   .(x, 1) -> x
   .(i(x), x) -> 1
   .(x, i(x)) -> 1
   .(i(y), .(y, z)) -> z
   .(y, .(i(y), z)) -> z
   .(.(x, y), z) -> .(x, .(y, z))

Q is empty.

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   .1(.(x, y), z) -> .1(y, z)
   .1(.(x, y), z) -> .1(x, .(y, z))
The following rules are removed from R:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(.(x_1, x_2)) = 2 + x_1 + x_2
   POL(.1(x_1, x_2)) = 2*x_1 + x_2
   POL(1) = 2
   POL(i(x_1)) = 1 + x_1


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

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

   .(.(x, y), z) -> .(x, .(y, z))

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

(10) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(11)
YES

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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(i, app(app(., x), y)) -> APP(i, x)
   APP(i, app(app(., x), y)) -> APP(i, y)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   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)

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

(13) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   APP(i, app(app(., x), y)) -> APP(i, x)
   APP(i, app(app(., x), y)) -> APP(i, y)

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

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

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

   i(.(x, y)) -> i(x)
   i(.(x, y)) -> i(y)

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

(17) 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:
*i(.(x, y)) -> i(x)
The graph contains the following edges 1 > 1


*i(.(x, y)) -> i(y)
The graph contains the following edges 1 > 1


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

(18)
YES

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

(19)
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(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   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)

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

(20) 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(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
The graph contains the following edges 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


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

(21)
YES