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 QDP
    ↳7 QDPOrderProof (⇔)
    ↳8 QDP
    ↳9 PisEmptyProof (⇔)
    ↳10 TRUE
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 DependencyGraphProof (⇔)
    ↳15 TRUE

(0) Obligation:

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

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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(., 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(i, app(app(., x), y)) → APP(app(., app(i, y)), app(i, x))
APP(i, app(app(., x), y)) → APP(., app(i, y))
APP(i, app(app(., x), y)) → APP(i, y)
APP(i, app(app(., x), y)) → APP(i, x)
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(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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 12 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(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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) Obligation:

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

APP(i, app(app(., x), y)) → APP(i, x)
APP(i, app(app(., x), y)) → APP(i, y)

The TRS R consists of the following rules:

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.

(7) 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

i1(.(x, y)) → i1(x)
i1(.(x, y)) → i1(y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(i, app(app(., x), y)) → APP(i, x)
APP(i, app(app(., x), y)) → APP(i, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
i1(x1)  =  x1
.(x1, x2)  =  .(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE

(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(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(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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(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
.  =  .
1  =  1
i  =  i
nil  =  nil

Recursive Path Order [RPO].
Precedence:
map > cons > [app2, true, 1, nil] > APP1
filter > filter2 > cons > [app2, true, 1, nil] > APP1
false > [app2, true, 1, nil] > APP1
i > . > [app2, true, 1, nil] > APP1


The following usable rules [FROCOS05] were oriented: none

(13) 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(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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) DependencyGraphProof (EQUIVALENT transformation)

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

(15) TRUE