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))


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 contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SUM2(cons2(0, x), y) -> SUM2(x, y)
Used argument filtering: SUM2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
              ↳ 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))

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SUM2(cons2(s1(n), x), cons2(m, y)) -> SUM2(cons2(n, x), cons2(s1(m), y))
Used argument filtering: SUM2(x1, x2)  =  x1
cons2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
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
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

WEIGHT1(cons2(n, cons2(m, x))) -> WEIGHT1(sum2(cons2(n, cons2(m, x)), cons2(0, x)))
Used argument filtering: WEIGHT1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
sum2(x1, x2)  =  x2
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
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.