Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
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(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(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(g, 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), z)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(:, app(app(+, x), y)), z) → APP(:, x)

The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
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(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(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(g, 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), z)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(:, app(app(+, x), y)), z) → APP(:, x)

The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
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(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(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, 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(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
APP(app(filter, 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(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(g, z)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(:, app(app(+, x), y)), z) → APP(:, x)

The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 20 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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(:, x), z)
APP(app(:, app(app(:, x), y)), z) → 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(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
app(x1, x2)  =  app(x1, x2)
:  =  :
+  =  +
g  =  g
a  =  a

Lexicographic path order with status [19].
Quasi-Precedence:
[:, g] > + > app2
a > app2

Status:
a: multiset
:: multiset
app2: [1,2]
+: multiset
g: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
filter2  =  filter2
true  =  true
filter  =  filter
map  =  map
cons  =  cons
false  =  false

Lexicographic path order with status [19].
Quasi-Precedence:
true > filter > [APP1, app2, filter2, cons]
map > [APP1, app2, filter2, cons]
false > filter > [APP1, app2, filter2, cons]

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


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof

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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

MAP(f, cons(x, xs)) → MAP(f, xs)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.


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.
none
Used ordering: Combined order from the following AFS and order.
MAP(x1, x2)  =  MAP(x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
cons2 > MAP1

Status:
MAP1: [1]
cons2: [1,2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R 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(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.