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:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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:

SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
DBL1(s1(X)) -> DBL1(X)

The TRS R consists of the following rules:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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:

SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
DBL1(s1(X)) -> DBL1(X)

The TRS R consists of the following rules:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

ADD2(s1(X), Y) -> ADD2(X, Y)

The TRS R consists of the following rules:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD2(x1, x2)) = 3·x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

DBL1(s1(X)) -> DBL1(X)

The TRS R consists of the following rules:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.


DBL1(s1(X)) -> DBL1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DBL1(x1)) = 3·x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

SQR1(s1(X)) -> SQR1(X)

The TRS R consists of the following rules:

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.


SQR1(s1(X)) -> SQR1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SQR1(x1)) = 3·x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

terms1(N) -> cons1(recip1(sqr1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons1(Y)) -> cons1(Y)

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.