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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(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:

SUM1(s(x)) → SUM1(x)
SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

SUM1(s(x)) → SUM1(x)
SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

SUM1(s(x)) → SUM1(x)
SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

SUM1(s(x)) → SUM1(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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.


SUM(s(x)) → SUM(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
sum1(0)
sum1(s(x0))

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