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:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

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:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SUM11(s1(x)) -> SUM11(x)
SUM1(s1(x)) -> SUM1(x)

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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:

SUM11(s1(x)) -> SUM11(x)
SUM1(s1(x)) -> SUM1(x)

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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

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

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

SUM11(s1(x)) -> SUM11(x)

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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.


SUM11(s1(x)) -> SUM11(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM11(x1)  =  SUM11(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > SUM11


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:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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:

SUM1(s1(x)) -> SUM1(x)

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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.


SUM1(s1(x)) -> SUM1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM1(x1)  =  SUM1(x1)
s1(x1)  =  s1(x1)

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


The following usable rules [14] were oriented: none



↳ 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:

sum1(0) -> 0
sum1(s1(x)) -> +2(sum1(x), s1(x))
sum11(0) -> 0
sum11(s1(x)) -> s1(+2(sum11(x), +2(x, x)))

The set Q consists of the following terms:

sum1(0)
sum1(s1(x0))
sum11(0)
sum11(s1(x0))

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