Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHExSec11_1_Luc02a

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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(X)

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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(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:

MARK₁(half₁(X)) -> A__HALF₁(mark₁(X))
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
A__SQR₁(s₁(X)) -> MARK₁(X)
MARK₁(sqr₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(dbl₁(X)) -> A__DBL₁(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)
A__SQR₁(s₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(terms₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(terms₁(X)) -> A__TERMS₁(mark₁(X))
A__ADD₂(s₁(X), Y) -> A__ADD₂(mark₁(X), mark₁(Y))
A__DBL₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(dbl₁(X)) -> MARK₁(X)
A__HALF₁(s₁(s₁(X))) -> A__HALF₁(mark₁(X))
MARK₁(sqr₁(X)) -> MARK₁(X)
A__SQR₁(s₁(X)) -> A__ADD₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X)))
MARK₁(half₁(X)) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> MARK₁(X)
A__HALF₁(dbl₁(X)) -> MARK₁(X)
A__TERMS₁(N) -> MARK₁(N)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
A__HALF₁(s₁(s₁(X))) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> MARK₁(Y)
A__DBL₁(s₁(X)) -> A__DBL₁(mark₁(X))
A__TERMS₁(N) -> A__SQR₁(mark₁(N))
A__SQR₁(s₁(X)) -> A__DBL₁(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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK₁(half₁(X)) -> A__HALF₁(mark₁(X))
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
A__SQR₁(s₁(X)) -> MARK₁(X)
MARK₁(sqr₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(dbl₁(X)) -> A__DBL₁(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)
A__SQR₁(s₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(terms₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(terms₁(X)) -> A__TERMS₁(mark₁(X))
A__ADD₂(s₁(X), Y) -> A__ADD₂(mark₁(X), mark₁(Y))
A__DBL₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(dbl₁(X)) -> MARK₁(X)
A__HALF₁(s₁(s₁(X))) -> A__HALF₁(mark₁(X))
MARK₁(sqr₁(X)) -> MARK₁(X)
A__SQR₁(s₁(X)) -> A__ADD₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X)))
MARK₁(half₁(X)) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> MARK₁(X)
A__HALF₁(dbl₁(X)) -> MARK₁(X)
A__TERMS₁(N) -> MARK₁(N)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
A__HALF₁(s₁(s₁(X))) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> MARK₁(Y)
A__DBL₁(s₁(X)) -> A__DBL₁(mark₁(X))
A__TERMS₁(N) -> A__SQR₁(mark₁(N))
A__SQR₁(s₁(X)) -> A__DBL₁(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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(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 strictly oriented and are deleted.

MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
A__SQR₁(s₁(X)) -> MARK₁(X)
MARK₁(dbl₁(X)) -> A__DBL₁(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)
A__SQR₁(s₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(terms₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> A__ADD₂(mark₁(X), mark₁(Y))
A__DBL₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(dbl₁(X)) -> MARK₁(X)
A__HALF₁(s₁(s₁(X))) -> A__HALF₁(mark₁(X))
MARK₁(sqr₁(X)) -> MARK₁(X)
A__SQR₁(s₁(X)) -> A__ADD₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X)))
A__ADD₂(s₁(X), Y) -> MARK₁(X)
A__HALF₁(dbl₁(X)) -> MARK₁(X)
A__TERMS₁(N) -> MARK₁(N)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
A__HALF₁(s₁(s₁(X))) -> MARK₁(X)
A__ADD₂(s₁(X), Y) -> MARK₁(Y)
A__DBL₁(s₁(X)) -> A__DBL₁(mark₁(X))
A__TERMS₁(N) -> A__SQR₁(mark₁(N))
A__SQR₁(s₁(X)) -> A__DBL₁(mark₁(X))

The remaining pairs can at least by weakly be oriented.

MARK₁(half₁(X)) -> A__HALF₁(mark₁(X))
MARK₁(sqr₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(terms₁(X)) -> A__TERMS₁(mark₁(X))
MARK₁(half₁(X)) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = x1
half₁(x1) = x1
A__HALF₁(x1) = x1
mark₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
A__FIRST₂(x1, x2) = A__FIRST₂(x1, x2)
A__SQR₁(x1) = A__SQR₁(x1)
s₁(x1) = s₁(x1)
sqr₁(x1) = sqr₁(x1)
dbl₁(x1) = dbl₁(x1)
A__DBL₁(x1) = A__DBL₁(x1)
add₂(x1, x2) = add₂(x1, x2)
recip₁(x1) = recip₁(x1)
cons₂(x1, x2) = cons₁(x1)
terms₁(x1) = terms₁(x1)
A__ADD₂(x1, x2) = A__ADD₂(x1, x2)
0 = 0
A__TERMS₁(x1) = A__TERMS₁(x1)
a__sqr₁(x1) = a__sqr₁(x1)
a__dbl₁(x1) = a__dbl₁(x1)
a__terms₁(x1) = a__terms₁(x1)
a__add₂(x1, x2) = a__add₂(x1, x2)
a__half₁(x1) = x1
a__first₂(x1, x2) = a__first₂(x1, x2)
nil = nil

Lexicographic Path Order [19].
Precedence:

[first₂, a_{_first₂}, nil] > A_{_FIRST₂} > [A_{_DBL₁}, cons_1]
[terms₁, A_{_TERMS₁}, a_{_{terms_1]}} > [A_{_SQR₁}, sqr₁, a_{_{sqr_1]}} > [dbl₁, a_{_{dbl_1]}} > s₁ > [A_{_DBL₁}, cons_1]
[terms₁, A_{_TERMS₁}, a_{_{terms_1]}} > [A_{_SQR₁}, sqr₁, a_{_{sqr_1]}} > [add₂, a_{_{add_2]}} > s₁ > [A_{_DBL₁}, cons_1]
[terms₁, A_{_TERMS₁}, a_{_{terms_1]}} > [A_{_SQR₁}, sqr₁, a_{_{sqr_1]}} > [add₂, a_{_{add_2]}} > A_{_ADD₂} > [A_{_DBL₁}, cons_1]
[terms₁, A_{_TERMS₁}, a_{_{terms_1]}} > [A_{_SQR₁}, sqr₁, a_{_{sqr_1]}} > 0 > [A_{_DBL₁}, cons_1]
[terms₁, A_{_TERMS₁}, a_{_{terms_1]}} > recip₁ > [A_{_DBL₁}, cons_1]

The following usable rules [14] were oriented:

mark₁(terms₁(X)) -> a__terms₁(mark₁(X))
mark₁(sqr₁(X)) -> a__sqr₁(mark₁(X))
a__add₂(0, X) -> mark₁(X)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(half₁(X)) -> a__half₁(mark₁(X))
a__half₁(dbl₁(X)) -> mark₁(X)
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(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__first₂(X1, X2) -> first₂(X1, X2)
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__dbl₁(X) -> dbl₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__add₂(X1, X2) -> add₂(X1, X2)
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(X) -> half₁(X)
a__sqr₁(0) -> 0
a__sqr₁(s₁(X)) -> s₁(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__sqr₁(X) -> sqr₁(X)
a__terms₁(N) -> cons₂(recip₁(a__sqr₁(mark₁(N))), terms₁(s₁(N)))
a__terms₁(X) -> terms₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK₁(half₁(X)) -> A__HALF₁(mark₁(X))
MARK₁(sqr₁(X)) -> A__SQR₁(mark₁(X))
MARK₁(terms₁(X)) -> A__TERMS₁(mark₁(X))
MARK₁(half₁(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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK₁(half₁(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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(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 strictly oriented and are deleted.

MARK₁(half₁(X)) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
half₁(x1) = half₁(x1)

Lexicographic Path Order [19].
Precedence:

half₁ > MARK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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(a__add₂(a__sqr₁(mark₁(X)), a__dbl₁(mark₁(X))))
a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(a__dbl₁(mark₁(X))))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(a__add₂(mark₁(X), mark₁(Y)))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__half₁(0) -> 0
a__half₁(s₁(0)) -> 0
a__half₁(s₁(s₁(X))) -> s₁(a__half₁(mark₁(X)))
a__half₁(dbl₁(X)) -> mark₁(X)
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(half₁(X)) -> a__half₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(recip₁(X)) -> recip₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(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)
a__half₁(X) -> half₁(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.