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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → 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(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(double, app(s, x)) → APP(s, app(s, app(double, 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(app(map, f), xs)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(plus, app(s, x)), y) → APP(s, y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
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(double, app(s, x)) → APP(double, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(plus, app(s, x)), y) → APP(double, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
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(plus, app(s, x)), y) → APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
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(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) → APP(minus, x)
APP(double, app(s, x)) → APP(s, app(double, x))
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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → 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(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(double, app(s, x)) → APP(s, app(s, app(double, 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(app(map, f), xs)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(plus, app(s, x)), y) → APP(s, y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
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(double, app(s, x)) → APP(double, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(plus, app(s, x)), y) → APP(double, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
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(plus, app(s, x)), y) → APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
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(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) → APP(minus, x)
APP(double, app(s, x)) → APP(s, app(double, x))
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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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 4 SCCs with 20 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP(double, app(s, x)) → APP(double, x)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ ATransformationProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP(double, app(s, x)) → APP(double, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ ATransformationProof
QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

double(s(x)) → double(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ ATransformationProof
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ ATransformationProof
QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

minus(s(x), s(y)) → minus(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPApplicativeOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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].Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

plus1(s(x), y) → plus1(x, y)
plus1(s(x), y) → plus1(x, s(y))
plus1(s(x), y) → plus1(minus(x, y), double(y))

The a-transformed usable rules are

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x


The following pairs can be oriented strictly and are deleted.


APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
The remaining pairs can at least be oriented weakly.

APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(double(x1)) = 0   
POL(minus(x1, x2)) = 1 + x1   
POL(plus1(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → x



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

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

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

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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].Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

plus1(s(x), y) → plus1(minus(x, y), double(y))

The a-transformed usable rules are

double(0) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x


The following pairs can be oriented strictly and are deleted.


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

POL(0) = 0   
POL(double(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(plus1(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

app(double, 0) → 0
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → x



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

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
We use the reduction pair processor [15]. Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

plus1(s(x), y) → plus1(minus(x, y), double(y))

The a-transformed usable rules are

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))


The following pairs can be oriented strictly and are deleted.


APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( minus(x1, x2) ) =
/0\
\0/
+
/10\
\01/
·x1+
/00\
\11/
·x2

M( s(x1) ) =
/1\
\1/
+
/10\
\01/
·x1

M( double(x1) ) =
/1\
\1/
+
/11\
\11/
·x1

M( 0 ) =
/0\
\0/

Tuple symbols:
M( plus1(x1, x2) ) = 0+
[1,0]
·x1+
[0,0]
·x2


Matrix type:
We used a basic matrix type which is not further parametrizeable.


The following usable rules [17] were oriented:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))



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

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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
QDP
            ↳ QDPSizeChangeProof

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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: