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:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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:

APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(size, app2(app2(node, x), xs)) -> APP2(app2(map, size), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(plus, x)
APP2(size, app2(app2(node, x), xs)) -> APP2(map, size)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(app2(plus, x), app2(sum, xs))
APP2(size, app2(app2(node, x), xs)) -> APP2(s, app2(sum, app2(app2(map, size), xs)))
APP2(size, app2(app2(node, x), xs)) -> APP2(sum, app2(app2(map, size), xs))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(sum, xs)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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:

APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(size, app2(app2(node, x), xs)) -> APP2(app2(map, size), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(plus, x)
APP2(size, app2(app2(node, x), xs)) -> APP2(map, size)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(app2(plus, x), app2(sum, xs))
APP2(size, app2(app2(node, x), xs)) -> APP2(s, app2(sum, app2(app2(map, size), xs)))
APP2(size, app2(app2(node, x), xs)) -> APP2(sum, app2(app2(map, size), xs))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(sum, app2(app2(cons, x), xs)) -> APP2(sum, xs)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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 9 less nodes.

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

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

APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)

The TRS R consists of the following rules:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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.


APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 1}


POL( app2(x1, x2) ) = x2 + 1



The following usable rules [14] 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:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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
            ↳ QDPOrderProof
          ↳ QDP

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

APP2(sum, app2(app2(cons, x), xs)) -> APP2(sum, xs)

The TRS R consists of the following rules:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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.


APP2(sum, app2(app2(cons, x), xs)) -> APP2(sum, xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 1}


POL( app2(x1, x2) ) = x1 + x2 + 1


POL( cons ) = 0



The following usable rules [14] were oriented: none



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

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

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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
            ↳ QDPOrderProof

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

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(size, app2(app2(node, x), xs)) -> APP2(app2(map, size), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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.


APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(size, app2(app2(node, x), xs)) -> APP2(app2(map, size), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 1}


POL( app2(x1, x2) ) = x1 + x2 + 1


POL( cons ) = 0


POL( node ) = 0



The following usable rules [14] were oriented: none



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

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

app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(sum, app2(app2(cons, x), xs)) -> app2(app2(plus, x), app2(sum, xs))
app2(size, app2(app2(node, x), xs)) -> app2(s, app2(sum, app2(app2(map, size), xs)))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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.