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:

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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:

WEIGHT1(cons2(n, cons2(m, x))) -> SUM2(cons2(n, cons2(m, x)), cons2(0, x))
WEIGHT1(cons2(n, cons2(m, x))) -> WEIGHT1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
SUM2(cons2(0, x), y) -> SUM2(x, y)
SUM2(cons2(s1(n), x), cons2(m, y)) -> SUM2(cons2(n, x), cons2(s1(m), y))

The TRS R consists of the following rules:

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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:

WEIGHT1(cons2(n, cons2(m, x))) -> SUM2(cons2(n, cons2(m, x)), cons2(0, x))
WEIGHT1(cons2(n, cons2(m, x))) -> WEIGHT1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
SUM2(cons2(0, x), y) -> SUM2(x, y)
SUM2(cons2(s1(n), x), cons2(m, y)) -> SUM2(cons2(n, x), cons2(s1(m), y))

The TRS R consists of the following rules:

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

SUM2(cons2(s1(n), x), cons2(m, y)) -> SUM2(cons2(n, x), cons2(s1(m), y))
SUM2(cons2(0, x), y) -> SUM2(x, y)

The TRS R consists of the following rules:

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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.


SUM2(cons2(s1(n), x), cons2(m, y)) -> SUM2(cons2(n, x), cons2(s1(m), y))
SUM2(cons2(0, x), y) -> SUM2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 2


POL( s1(x1) ) = x1 + 3


POL( SUM2(x1, x2) ) = max{0, 3x1 - 2}


POL( cons2(x1, x2) ) = 3x1 + x2 + 3



The following usable rules [14] were oriented: none



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

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

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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

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

WEIGHT1(cons2(n, cons2(m, x))) -> WEIGHT1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))

The TRS R consists of the following rules:

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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.


WEIGHT1(cons2(n, cons2(m, x))) -> WEIGHT1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = max{0, -2}


POL( s1(x1) ) = max{0, -3}


POL( WEIGHT1(x1) ) = max{0, 2x1 - 3}


POL( sum2(x1, x2) ) = x2


POL( nil ) = 1


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



The following usable rules [14] were oriented:

sum2(nil, y) -> y
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))



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

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

sum2(cons2(s1(n), x), cons2(m, y)) -> sum2(cons2(n, x), cons2(s1(m), y))
sum2(cons2(0, x), y) -> sum2(x, y)
sum2(nil, y) -> y
weight1(cons2(n, cons2(m, x))) -> weight1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
weight1(cons2(n, nil)) -> n

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.