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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE

(0) Obligation:

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

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, 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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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 2 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Quasi-Precedence:
[app2, :]

Status:
app2: [1,2]
:: multiset
C: multiset


The following usable rules [FROCOS05] were oriented:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))

(7) Obligation:

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

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false

Recursive path order with status [RPO].
Quasi-Precedence:
[map, cons] > APP1 > app2
filter2 > APP1 > app2
true > filter > APP1 > app2
false > filter > APP1 > app2

Status:
APP1: [1]
app2: multiset
map: multiset
cons: multiset
filter: multiset
filter2: multiset
true: multiset
false: multiset


The following usable rules [FROCOS05] were oriented: none

(12) 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)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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].


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
cons > [APP2, app2, map]

Status:
APP2: [1,2]
app2: multiset
map: multiset
cons: multiset


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE