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(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
SUM1(app2(l, cons2(x, cons2(y, k)))) -> APP2(l, sum1(cons2(x, cons2(y, k))))
SUM1(cons2(x, cons2(y, l))) -> PLUS2(x, y)
SUM1(plus2(cons2(0, x), cons2(y, l))) -> SUM1(cons2(s1(x), cons2(y, l)))
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
SUM1(plus2(cons2(0, x), cons2(y, l))) -> PRED1(sum1(cons2(s1(x), cons2(y, l))))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(cons2(x, cons2(y, k)))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
APP2(cons2(x, l), k) -> APP2(l, k)

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
SUM1(app2(l, cons2(x, cons2(y, k)))) -> APP2(l, sum1(cons2(x, cons2(y, k))))
SUM1(cons2(x, cons2(y, l))) -> PLUS2(x, y)
SUM1(plus2(cons2(0, x), cons2(y, l))) -> SUM1(cons2(s1(x), cons2(y, l)))
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
SUM1(plus2(cons2(0, x), cons2(y, l))) -> PRED1(sum1(cons2(s1(x), cons2(y, l))))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(cons2(x, cons2(y, k)))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
APP2(cons2(x, l), k) -> APP2(l, k)

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

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

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

PLUS2(s1(x), y) -> PLUS2(x, y)

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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.


PLUS2(s1(x), y) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( PLUS2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



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

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

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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
          ↳ QDP

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

APP2(cons2(x, l), k) -> APP2(l, k)

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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(cons2(x, l), k) -> APP2(l, k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = x1 + x2 + 1


POL( APP2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



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

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

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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
          ↳ QDP

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

SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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.


SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1


POL( s1(x1) ) = 1


POL( SUM1(x1) ) = max{0, x1 - 1}


POL( cons2(x1, x2) ) = x2 + 1


POL( plus2(x1, x2) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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
          ↳ QDP
            ↳ QDPOrderProof

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

SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))

The TRS R consists of the following rules:

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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.


SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( SUM1(x1) ) = x1 + 1


POL( plus2(x1, x2) ) = max{0, x1 - 1}


POL( sum1(x1) ) = 1


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


POL( s1(x1) ) = x1


POL( 0 ) = max{0, -1}


POL( nil ) = max{0, -1}


POL( cons2(x1, x2) ) = x2 + 1



The following usable rules [14] were oriented:

app2(l, nil) -> l
sum1(cons2(x, nil)) -> cons2(x, nil)
app2(nil, k) -> k
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))



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

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

app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)

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.