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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.


↳ QTRS
  ↳ DependencyPairsProof

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.

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(f, app(s, x)) → APP(f, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
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(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(g, x), app(c, y)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(s, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(g, x), app(c, y)) → APP(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y))
APP(app(g, x), app(c, y)) → APP(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
APP(app(g, x), app(c, y)) → APP(if, app(f, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, x), y))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, app(s, x)), y))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(g, app(s, x))
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(g, x), app(c, y)) → APP(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

APP(f, app(s, x)) → APP(f, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
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(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(g, x), app(c, y)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(s, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(g, x), app(c, y)) → APP(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y))
APP(app(g, x), app(c, y)) → APP(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
APP(app(g, x), app(c, y)) → APP(if, app(f, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, x), y))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, app(s, x)), y))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(g, app(s, x))
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(g, x), app(c, y)) → APP(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
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(f, app(s, x)) → APP(f, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
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(g, x), app(c, y)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(s, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(g, x), app(c, y)) → APP(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
APP(app(g, x), app(c, y)) → APP(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y))
APP(app(g, x), app(c, y)) → APP(if, app(f, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, x), y))
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, app(s, x)), y))
APP(app(filter, 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(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(g, x), app(c, y)) → APP(g, app(s, x))
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(g, x), app(c, y)) → APP(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y)))
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 18 less nodes.

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

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, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
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:

F(S(x)) → F(x)

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(f, app(s, x)) → APP(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
S(x1)  =  S(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
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:

G(x, c(y)) → G(x, y)
G(x, c(y)) → G(s(x), y)

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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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(map, fun), app(app(cons, x), xs)) → APP(fun, x)
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, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
We use the reduction pair processor [13].


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

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)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, 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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.