Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
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(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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
      ↳ DependencyGraphProof

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

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
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(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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 [15,17,22] contains 3 SCCs with 14 less nodes.

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

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

APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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 [15].


The following pairs can be oriented strictly and are deleted.


APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (1/4)x_2   
POL(minus) = 0   
POL(app(x1, x2)) = (1/4)x_1 + (4)x_2   
POL(s) = 1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



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

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

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

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)
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(app(app(filter2, app(f, x)), f), x), xs)
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)

The TRS R consists of the following rules:

app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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 [15].


The following pairs can be oriented strictly and are deleted.


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

APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
Used ordering: Polynomial interpretation [25,35]:

POL(minus) = 0   
POL(true) = 0   
POL(pred) = 0   
POL(s) = 0   
POL(0) = 0   
POL(filter) = 0   
POL(APP(x1, x2)) = (1/4)x_2   
POL(quot) = 0   
POL(cons) = 0   
POL(map) = 0   
POL(false) = 0   
POL(app(x1, x2)) = 1/2 + (1/4)x_1 + (4)x_2   
POL(filter2) = 0   
POL(nil) = 0   
The value of delta used in the strict ordering is 5/32.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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), 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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
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 [15,17,22] contains 0 SCCs with 2 less nodes.