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 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔)
        ↳24 TRUE

(0) Obligation:

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

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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) 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(f, app(s, x)) → APP(f, x)
APP(g, app(app(cons, 0), y)) → APP(g, y)
APP(h, app(app(cons, x), y)) → APP(h, app(g, app(app(cons, x), y)))
APP(h, app(app(cons, x), y)) → APP(g, app(app(cons, x), y))
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(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(g, app(app(cons, 0), y)) → APP(g, y)

The TRS R consists of the following rules:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
g: []
app2: [2,1]
cons: []
0: []


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(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
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(h, app(app(cons, x), y)) → APP(h, app(g, app(app(cons, x), y)))

The TRS R consists of the following rules:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
cons > [g, s] > h
0 > h

Status:
h: []
cons: []
g: []
0: []
s: []


The following usable rules [FROCOS05] were oriented:

app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)

(12) Obligation:

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

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

APP(f, app(s, x)) → APP(f, x)

The TRS R consists of the following rules:

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(f, app(s, x)) → 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)
f  =  f
app(x1, x2)  =  app(x1, x2)
s  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
APP1 > f
app2 > f
s > f

Status:
APP1: [1]
f: []
app2: [1,2]
s: []


The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)

The TRS R consists of the following rules:

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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(fun, 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

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

Status:
APP1: [1]
app2: [2,1]
map: []
cons: []
filter: []
filter2: []
true: []
false: []


The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)

The TRS R consists of the following rules:

app(f, app(s, x)) → app(f, x)
app(g, app(app(cons, 0), y)) → app(g, y)
app(g, app(app(cons, app(s, x)), y)) → app(s, x)
app(h, app(app(cons, x), y)) → app(h, app(g, app(app(cons, x), y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) TRUE