(0) Obligation:

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

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, app(s, z)), 0)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(times, app(s, z))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
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(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

plus1(x, s(y)) → plus1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:
s1 > plus12

Status:
s1: multiset
plus12: [1,2]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

times1(x, plus(y, s(z))) → times1(x, s(z))
times1(x, s(y)) → times1(x, y)
times1(x, plus(y, s(z))) → times1(x, plus(y, times(s(z), 0)))

The a-transformed usable rules are

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))


The following pairs can be oriented strictly and are deleted.


APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2)  =  x2
plus(x1, x2)  =  plus(x1, x2)
s(x1)  =  x1
times(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Precedence:
plus2 > 0

Status:
plus2: [1,2]
0: multiset

The following usable rules [FROCOS05] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

times1(x, s(y)) → times1(x, y)
times1(x, plus(y, s(z))) → times1(x, plus(y, times(s(z), 0)))

The a-transformed usable rules are

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))


The following pairs can be oriented strictly and are deleted.


APP(app(times, x), app(s, y)) → APP(app(times, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2)  =  times1(x2)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Precedence:
times11 > s1 > 0
plus2 > s1 > 0

Status:
plus2: multiset
times11: multiset
s1: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

times1(x, plus(y, s(z))) → times1(x, plus(y, times(s(z), 0)))

The a-transformed usable rules are

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))


The following pairs can be oriented strictly and are deleted.


APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2)  =  x2
plus(x1, x2)  =  plus(x1, x2)
s(x1)  =  s
times(x1, x2)  =  x2
0  =  0

Recursive path order with status [RPO].
Precedence:
plus2 > s > 0

Status:
plus2: multiset
s: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))

(16) Obligation:

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

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) TRUE

(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(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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(filter, f), app(app(cons, x), xs)) → APP(f, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
times  =  times
0  =  0
plus  =  plus
s  =  s
nil  =  nil

Recursive path order with status [RPO].
Precedence:
filter2 > cons > map > app2 > APP1
true > cons > map > app2 > APP1
true > filter > app2 > APP1
false > app2 > APP1
0 > APP1
plus > times > s > app2 > APP1
nil > APP1

Status:
APP1: multiset
plus: multiset
true: multiset
s: multiset
0: multiset
filter: multiset
cons: multiset
map: multiset
times: multiset
false: multiset
app2: multiset
filter2: multiset
nil: multiset

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, 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)

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

(22) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(23) TRUE