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(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

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:

SUM1(s1(x)) -> SUM1(x)
SUM1(s1(x)) -> SQR1(s1(x))

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

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:

SUM1(s1(x)) -> SUM1(x)
SUM1(s1(x)) -> SQR1(s1(x))

The TRS R consists of the following rules:

sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

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.


SUM1(s1(x)) -> SUM1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SUM1(x1)) = 3·x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))

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.