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(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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) 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(g, app(g, x)) → APP(g, app(h, app(g, x)))
APP(g, app(g, x)) → APP(h, app(g, x))
APP(h, app(h, x)) → APP(h, app(app(f, app(h, x)), x))
APP(h, app(h, x)) → APP(app(f, app(h, x)), x)
APP(h, app(h, x)) → APP(f, app(h, 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(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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(g, app(g, x)) → APP(g, app(h, app(g, x)))

The TRS R consists of the following rules:

app(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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

g1(g(x)) → g1(h(g(x)))

The a-transformed usable rules are

g(g(x)) → g(h(g(x)))
g(h(g(x))) → g(x)
h(h(x)) → h(f(h(x), x))


The following pairs can be oriented strictly and are deleted.


APP(g, app(g, x)) → APP(g, app(h, app(g, x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
g1(x1)  =  g1(x1)
g(x1)  =  g
h(x1)  =  h
f(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
g > [g11, h]


The following usable rules [FROCOS05] were oriented:

app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(g, app(h, app(g, x))) → app(g, x)
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), x))

(7) Obligation:

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

app(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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, 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(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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, 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 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 [RPO].
Precedence:
map > [APP1, app2] > [filter, true]
filter2 > [APP1, app2] > [filter, true]
false > [filter, true]


The following usable rules [FROCOS05] were oriented: none

(12) 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)

The TRS R consists of the following rules:

app(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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

map1(fun, cons(x, xs)) → map1(fun, xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


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(g, app(h, app(g, x))) → app(g, x)
app(g, app(g, x)) → app(g, app(h, app(g, x)))
app(h, app(h, x)) → app(h, app(app(f, app(h, x)), 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.
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