Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

app(app(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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(*, x), app(app(+, y), z)) → APP(app(+, app(app(*, x), y)), app(app(*, x), z))
APP(app(*, x), app(app(+, y), z)) → APP(+, app(app(*, x), y))
APP(app(*, x), app(app(+, y), z)) → APP(app(*, x), y)
APP(app(*, x), app(app(+, y), z)) → APP(app(*, x), z)
APP(app(*, app(app(+, y), z)), x) → APP(app(+, app(app(*, x), y)), app(app(*, x), z))
APP(app(*, app(app(+, y), z)), x) → APP(+, app(app(*, x), y))
APP(app(*, app(app(+, y), z)), x) → APP(app(*, x), y)
APP(app(*, app(app(+, y), z)), x) → APP(*, x)
APP(app(*, app(app(+, y), z)), x) → APP(app(*, x), z)
APP(app(*, app(app(*, x), y)), z) → APP(app(*, x), app(app(*, y), z))
APP(app(*, app(app(*, x), y)), z) → APP(app(*, y), z)
APP(app(*, app(app(*, x), y)), z) → APP(*, y)
APP(app(+, app(app(+, x), y)), z) → APP(app(+, x), app(app(+, y), z))
APP(app(+, app(app(+, x), y)), z) → APP(app(+, y), z)
APP(app(+, app(app(+, x), y)), z) → APP(+, 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(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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(+, app(app(+, x), y)), z) → APP(app(+, y), z)
APP(app(+, app(app(+, x), y)), z) → APP(app(+, x), app(app(+, y), z))

The TRS R consists of the following rules:

app(app(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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

+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))

The a-transformed usable rules are

+(+(x, y), z) → +(x, +(y, z))


The following pairs can be oriented strictly and are deleted.


APP(app(+, app(app(+, x), y)), z) → APP(app(+, y), z)
APP(app(+, app(app(+, x), y)), z) → APP(app(+, x), app(app(+, y), z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1)
+(x1, x2)  =  +(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
[+11, +2]


The following usable rules [FROCOS05] were oriented:

app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))

(7) Obligation:

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

app(app(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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(*, x), app(app(+, y), z)) → APP(app(*, x), z)
APP(app(*, x), app(app(+, y), z)) → APP(app(*, x), y)
APP(app(*, app(app(+, y), z)), x) → APP(app(*, x), y)
APP(app(*, app(app(+, y), z)), x) → APP(app(*, x), z)
APP(app(*, app(app(*, x), y)), z) → APP(app(*, x), app(app(*, y), z))
APP(app(*, app(app(*, x), y)), z) → APP(app(*, y), z)

The TRS R consists of the following rules:

app(app(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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) 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(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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.

(12) 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)  =  x1
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false

Lexicographic Path Order [LPO].
Precedence:
[map, cons] > [app2, filter, false]
filter2 > [app2, filter, false]
true > [app2, filter, false]


The following usable rules [FROCOS05] were oriented: none

(13) 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(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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.

(14) 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(f, cons(x, xs)) → map1(f, xs)

The a-transformed usable rules are
none


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: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

app(app(*, x), app(app(+, y), z)) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(+, y), z)), x) → app(app(+, app(app(*, x), y)), app(app(*, x), z))
app(app(*, app(app(*, x), y)), z) → app(app(*, x), app(app(*, y), z))
app(app(+, app(app(+, x), y)), z) → app(app(+, x), app(app(+, y), z))
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.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE