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
  ↳ Non-Overlap Check

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.

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:

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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))


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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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:

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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD2(x1, x2)  =  ADD1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[ADD1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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
              ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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.


DBL1(s1(X)) -> DBL1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  DBL1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > DBL1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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
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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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.


SQR1(s1(X)) -> SQR1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SQR1(x1)  =  SQR1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > SQR1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons1(x1))

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