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:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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:

MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(dbl1(X)) -> MARK1(X)
MARK1(sqr1(X)) -> A__SQR1(mark1(X))
MARK1(dbl1(X)) -> A__DBL1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(recip1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(terms1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X1)
A__TERMS1(N) -> A__SQR1(mark1(N))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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:

MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(dbl1(X)) -> MARK1(X)
MARK1(sqr1(X)) -> A__SQR1(mark1(X))
MARK1(dbl1(X)) -> A__DBL1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(recip1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(terms1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X1)
A__TERMS1(N) -> A__SQR1(mark1(N))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

MARK1(dbl1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(sqr1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(recip1(X)) -> MARK1(X)
MARK1(terms1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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.


MARK1(dbl1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
The remaining pairs can at least be oriented weakly.

MARK1(sqr1(X)) -> MARK1(X)
MARK1(recip1(X)) -> MARK1(X)
MARK1(terms1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark1(x1) ) = x1


POL( dbl1(x1) ) = 2x1 + 3


POL( first2(x1, x2) ) = 3x1 + x2 + 1


POL( a__first2(x1, x2) ) = 3x1 + x2 + 1


POL( MARK1(x1) ) = x1 + 1


POL( 0 ) = 1


POL( A__ADD2(x1, x2) ) = x1 + x2 + 3


POL( a__add2(x1, x2) ) = x1 + x2 + 3


POL( A__TERMS1(x1) ) = 2x1 + 1


POL( nil ) = 1


POL( cons2(x1, x2) ) = 2x1


POL( a__dbl1(x1) ) = 2x1 + 3


POL( a__terms1(x1) ) = 2x1


POL( sqr1(x1) ) = x1


POL( recip1(x1) ) = x1


POL( terms1(x1) ) = 2x1


POL( add2(x1, x2) ) = x1 + x2 + 3


POL( s1(x1) ) = max{0, -3}


POL( A__FIRST2(x1, x2) ) = 3x1 + x2 + 1


POL( a__sqr1(x1) ) = x1



The following usable rules [14] were oriented:

a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__sqr1(0) -> 0
a__add2(X1, X2) -> add2(X1, X2)
mark1(s1(X)) -> s1(X)
a__add2(0, X) -> mark1(X)
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
a__first2(0, X) -> nil
a__dbl1(X) -> dbl1(X)
mark1(0) -> 0
mark1(recip1(X)) -> recip1(mark1(X))
a__sqr1(X) -> sqr1(X)
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__first2(X1, X2) -> first2(X1, X2)
mark1(dbl1(X)) -> a__dbl1(mark1(X))
a__terms1(X) -> terms1(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(nil) -> nil
a__dbl1(0) -> 0
mark1(sqr1(X)) -> a__sqr1(mark1(X))
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))



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

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

A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
MARK1(recip1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(terms1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPOrderProof

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

A__TERMS1(N) -> MARK1(N)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
MARK1(recip1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(terms1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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.


MARK1(terms1(X)) -> A__TERMS1(mark1(X))
MARK1(recip1(X)) -> MARK1(X)
MARK1(terms1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

A__TERMS1(N) -> MARK1(N)
MARK1(sqr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark1(x1) ) = x1


POL( dbl1(x1) ) = 3


POL( first2(x1, x2) ) = 2x2 + 1


POL( a__first2(x1, x2) ) = 2x2 + 1


POL( MARK1(x1) ) = 2x1


POL( 0 ) = max{0, -3}


POL( a__add2(x1, x2) ) = x1 + x2


POL( A__TERMS1(x1) ) = 2x1


POL( nil ) = max{0, -3}


POL( cons2(x1, x2) ) = x1


POL( a__dbl1(x1) ) = 3


POL( a__terms1(x1) ) = 2x1 + 3


POL( recip1(x1) ) = 2x1 + 2


POL( sqr1(x1) ) = x1


POL( terms1(x1) ) = 2x1 + 3


POL( add2(x1, x2) ) = x1 + x2


POL( s1(x1) ) = 2


POL( a__sqr1(x1) ) = x1



The following usable rules [14] were oriented:

a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__sqr1(0) -> 0
a__add2(X1, X2) -> add2(X1, X2)
mark1(s1(X)) -> s1(X)
a__add2(0, X) -> mark1(X)
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
a__first2(0, X) -> nil
a__dbl1(X) -> dbl1(X)
mark1(0) -> 0
mark1(recip1(X)) -> recip1(mark1(X))
a__sqr1(X) -> sqr1(X)
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__first2(X1, X2) -> first2(X1, X2)
mark1(dbl1(X)) -> a__dbl1(mark1(X))
a__terms1(X) -> terms1(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(nil) -> nil
a__dbl1(0) -> 0
mark1(sqr1(X)) -> a__sqr1(mark1(X))
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))



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

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

A__TERMS1(N) -> MARK1(N)
MARK1(sqr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
QDP
                          ↳ QDPOrderProof

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

MARK1(sqr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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.


MARK1(sqr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1 + 2


POL( cons2(x1, x2) ) = 2x1 + 2


POL( sqr1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(dbl1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(X)
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)

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.