Termination w.r.t. Q proof of TRS/TRCSREx26_Luc03b

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__ADD₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(A__FIRST₂(x₁, x₂)) = 2 + 2·x₂
POL(A__TERMS₁(x₁)) = 1 + 2·x₁
POL(MARK₁(x₁)) = 1 + 2·x₁
POL(a__add₂(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(a__dbl₁(x₁)) = 1 + 2·x₁
POL(a__first₂(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(a__sqr₁(x₁)) = x₁
POL(a__terms₁(x₁)) = 1 + 2·x₁
POL(add₂(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = x₁
POL(dbl₁(x₁)) = 1 + 2·x₁
POL(first₂(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(mark₁(x₁)) = 1 + 2·x₁
POL(nil) = 0
POL(recip₁(x₁)) = x₁
POL(s₁(x₁)) = 0
POL(sqr₁(x₁)) = x₁
POL(terms₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__ADD₂(0, X) -> MARK₁(X)
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)

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)

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)

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 interpretation [21]:

POL(0) = 1
POL(A__TERMS₁(x₁)) = 1 + 2·x₁
POL(MARK₁(x₁)) = 1 + 2·x₁
POL(a__add₂(x₁, x₂)) = 2 + 2·x₂
POL(a__dbl₁(x₁)) = 2
POL(a__first₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a__sqr₁(x₁)) = x₁
POL(a__terms₁(x₁)) = 2 + 2·x₁
POL(add₂(x₁, x₂)) = 2 + 2·x₂
POL(cons₂(x₁, x₂)) = x₁
POL(dbl₁(x₁)) = 2
POL(first₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(mark₁(x₁)) = 2·x₁
POL(nil) = 0
POL(recip₁(x₁)) = 2 + x₁
POL(s₁(x₁)) = 2
POL(sqr₁(x₁)) = x₁
POL(terms₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

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

↳ 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 interpretation [21]:

POL(MARK₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 2 + 2·x₁
POL(sqr₁(x₁)) = 2 + 2·x₁

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.