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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))


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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

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
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM2(x1, x2)  =  SUM1(x1)
cons2(x1, x2)  =  cons2(x1, x2)
s1(x1)  =  s1(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
cons2 > [SUM1, s1]
0 > [SUM1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
WEIGHT1(x1)  =  WEIGHT1(x1)
cons2(x1, x2)  =  cons2(x1, x2)
sum2(x1, x2)  =  sum1(x2)
0  =  0
s1(x1)  =  s
nil  =  nil

Lexicographic Path Order [19].
Precedence:
cons2 > WEIGHT1 > [0, s]
cons2 > sum1 > [0, s]
nil > [0, s]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

sum2(cons2(s1(x0), x1), cons2(x2, x3))
sum2(cons2(0, x0), x1)
sum2(nil, x0)
weight1(cons2(x0, cons2(x1, x2)))
weight1(cons2(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.