/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 12 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 2 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) QDPOrderProof [EQUIVALENT, 31 ms]
        (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) PisEmptyProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), 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(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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)


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

(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(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
   APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x)
   APP(f, 0) -> APP(s, 0)
   APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
   APP(f, app(s, x)) -> APP(minus, app(s, x))
   APP(f, app(s, x)) -> APP(g, app(f, x))
   APP(f, app(s, x)) -> APP(f, x)
   APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
   APP(g, app(s, x)) -> APP(minus, app(s, x))
   APP(g, app(s, x)) -> APP(f, app(g, x))
   APP(g, app(s, x)) -> APP(g, x)
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs))
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x))
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x)
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x))
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x)
   APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs))
   APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x)
   APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs)
   APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun)
   APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs)
   APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun)

The TRS R consists of the following rules:

   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 15 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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) 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(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)

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

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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.
----------------------------------------

(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:

   minus1(s(x), s(y)) -> minus1(x, y)

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

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))
   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.
----------------------------------------

(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].

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))
   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)


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

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

   minus1(s(x), s(y)) -> minus1(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:
*minus1(s(x), s(y)) -> minus1(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(f, app(s, x)) -> APP(g, app(f, x))
   APP(g, app(s, x)) -> APP(f, app(g, x))
   APP(f, app(s, x)) -> APP(f, x)
   APP(g, app(s, x)) -> APP(g, x)

The TRS R consists of the following rules:

   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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) 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(f, app(s, x)) -> APP(g, app(f, x))
   APP(g, app(s, x)) -> APP(f, app(g, x))
   APP(f, app(s, x)) -> APP(f, x)
   APP(g, app(s, x)) -> APP(g, x)

The TRS R consists of the following rules:

   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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.
----------------------------------------

(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:

   f1(s(x)) -> g1(f(x))
   g1(s(x)) -> f1(g(x))
   f1(s(x)) -> f1(x)
   g1(s(x)) -> g1(x)

The TRS R consists of the following rules:

   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))
   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.
----------------------------------------

(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].

   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)


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

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

   f1(s(x)) -> g1(f(x))
   g1(s(x)) -> f1(g(x))
   f1(s(x)) -> f1(x)
   g1(s(x)) -> g1(x)

The TRS R consists of the following rules:

   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

(23) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   f1(s(x)) -> f1(x)
   g1(s(x)) -> g1(x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
f1(x1)  =  f1(x1)

s(x1)  =  s(x1)

g1(x1)  =  g1(x1)

f(x1)  =  f(x1)

g(x1)  =  g(x1)

0  =  0

minus(x1, x2)  =  minus(x1)


Recursive path order with status [RPO].
Quasi-Precedence: [f1_1, g1_1] > [s_1, f_1, g_1, minus_1]
0 > [s_1, f_1, g_1, minus_1]

Status: f1_1: multiset status
s_1: [1]
g1_1: multiset status
f_1: [1]
g_1: [1]
0: multiset status
minus_1: [1]


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   f1(s(x)) -> g1(f(x))
   g1(s(x)) -> f1(g(x))

The TRS R consists of the following rules:

   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

(25) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   f1(s(x)) -> g1(f(x))
   g1(s(x)) -> f1(g(x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
f1(x1)  =  f1(x1)

s(x1)  =  s(x1)

g1(x1)  =  x1

f(x1)  =  f(x1)

g(x1)  =  x1

0  =  0

minus(x1, x2)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=3
   0=1
   f1_1=2
   f_1=4

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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

(26)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
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(27) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(28)
YES

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(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   app(app(minus, x), 0) -> x
   app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
   app(f, 0) -> app(s, 0)
   app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
   app(g, 0) -> 0
   app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(minus, x0), 0)
   app(app(minus, app(s, x0)), app(s, x1))
   app(f, 0)
   app(f, app(s, x0))
   app(g, 0)
   app(g, app(s, x0))
   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.
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(30) 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, fun), app(app(cons, x), xs)) -> APP(fun, x)
The graph contains the following edges 1 > 1, 2 > 2


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


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


*APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
The graph contains the following edges 2 > 2


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


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


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(31)
YES