Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

A__TERMS(N) → A__SQR(mark(N))
A__TERMS(N) → MARK(N)
A__SQR(s(X)) → A__ADD(a__sqr(mark(X)), a__dbl(mark(X)))
A__SQR(s(X)) → A__SQR(mark(X))
A__SQR(s(X)) → MARK(X)
A__SQR(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
A__ADD(s(X), Y) → MARK(X)
A__ADD(s(X), Y) → MARK(Y)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__TERMS(N) → MARK(N)
A__SQR(s(X)) → A__ADD(a__sqr(mark(X)), a__dbl(mark(X)))
A__SQR(s(X)) → A__SQR(mark(X))
A__SQR(s(X)) → MARK(X)
A__SQR(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
A__ADD(s(X), Y) → MARK(X)
A__ADD(s(X), Y) → MARK(Y)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__TERMS(x1)  =  A__TERMS(x1)
A__SQR(x1)  =  A__SQR(x1)
mark(x1)  =  x1
MARK(x1)  =  MARK(x1)
s(x1)  =  s(x1)
A__ADD(x1, x2)  =  A__ADD(x1, x2)
a__sqr(x1)  =  a__sqr(x1)
a__dbl(x1)  =  a__dbl(x1)
A__DBL(x1)  =  A__DBL(x1)
0  =  0
A__FIRST(x1, x2)  =  A__FIRST(x1, x2)
cons(x1, x2)  =  x1
terms(x1)  =  terms(x1)
sqr(x1)  =  sqr(x1)
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first(x1, x2)
recip(x1)  =  x1
a__terms(x1)  =  a__terms(x1)
a__add(x1, x2)  =  a__add(x1, x2)
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:

[A_TERMS1, A_SQR1, a_sqr1, terms₁, sqr₁, a_terms1] > [MARK₁, A_ADD2, A_DBL1, A_FIRST2]
[A_TERMS1, A_SQR1, a_sqr1, terms₁, sqr₁, a_terms1] > [a_dbl1, dbl₁] > s₁ > [first₂, a_first2]
[A_TERMS1, A_SQR1, a_sqr1, terms₁, sqr₁, a_terms1] > [a_dbl1, dbl₁] > 0 > nil
[A_TERMS1, A_SQR1, a_sqr1, terms₁, sqr₁, a_terms1] > [add₂, a_add2] > s₁ > [first₂, a_first2]

Status:

A_TERMS1: [1]
A_SQR1: [1]
MARK₁: multiset
s₁: multiset
A_ADD2: multiset
a_sqr1: [1]
a_dbl1: multiset
A_DBL1: multiset
0: multiset
A_FIRST2: multiset
terms₁: [1]
sqr₁: [1]
add₂: multiset
dbl₁: multiset
first₂: multiset
a_terms1: [1]
a_add2: multiset
a_first2: multiset
nil: multiset

The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(4) Obligation:

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

A__TERMS(N) → A__SQR(mark(N))
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

MARK(recip(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(recip(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
recip(x1)  =  recip(x1)
cons(x1, x2)  =  x1
a__terms(x1)  =  a__terms
a__sqr(x1)  =  a__sqr
mark(x1)  =  mark(x1)
terms(x1)  =  terms
s(x1)  =  s
0  =  0
a__add(x1, x2)  =  a__add(x1, x2)
a__dbl(x1)  =  a__dbl(x1)
a__first(x1, x2)  =  a__first(x2)
nil  =  nil
first(x1, x2)  =  first(x2)
sqr(x1)  =  sqr
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)

Recursive path order with status [RPO].
Quasi-Precedence:

[a_terms, terms] > [a_sqr, mark₁, a_add2, a_dbl1, a_first1, first₁, add₂, dbl₁] > recip₁
[a_terms, terms] > [a_sqr, mark₁, a_add2, a_dbl1, a_first1, first₁, add₂, dbl₁] > s
[a_terms, terms] > [a_sqr, mark₁, a_add2, a_dbl1, a_first1, first₁, add₂, dbl₁] > 0 > nil
[a_terms, terms] > [a_sqr, mark₁, a_add2, a_dbl1, a_first1, first₁, add₂, dbl₁] > sqr

Status:

MARK₁: [1]
recip₁: [1]
a_terms: []
a_sqr: []
mark₁: [1]
terms: []
s: multiset
0: multiset
a_add2: [2,1]
a_dbl1: [1]
a_first1: [1]
nil: multiset
first₁: [1]
sqr: multiset
add₂: [2,1]
dbl₁: [1]

The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(8) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
cons(x1, x2)  =  cons(x1)
a__terms(x1)  =  a__terms
recip(x1)  =  recip
a__sqr(x1)  =  x1
mark(x1)  =  mark(x1)
terms(x1)  =  terms
s(x1)  =  s
0  =  0
a__add(x1, x2)  =  a__add(x2)
a__dbl(x1)  =  x1
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil
first(x1, x2)  =  first(x1, x2)
sqr(x1)  =  x1
add(x1, x2)  =  add(x2)
dbl(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:

[a_terms, terms] > [mark₁, s, a_add1, a_first2, first₂, add₁] > cons₁
[a_terms, terms] > [mark₁, s, a_add1, a_first2, first₂, add₁] > recip
[0, nil] > [mark₁, s, a_add1, a_first2, first₂, add₁] > cons₁
[0, nil] > [mark₁, s, a_add1, a_first2, first₂, add₁] > recip

Status:

cons₁: multiset
a_terms: []
recip: multiset
mark₁: multiset
terms: []
s: multiset
0: multiset
a_add1: multiset
a_first2: multiset
nil: multiset
first₂: multiset
add₁: multiset

The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(10) Obligation:

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

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE