Termination w.r.t. Q proof of /tmp/tmpQnqCLK/Ex26_Luc03b

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 QDPOrderProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 PisEmptyProof (⇔)

↳16 TRUE

(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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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__ADD(0, X) → MARK(X)
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)

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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

A__TERMS(N) → MARK(N)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(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)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
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.
A__TERMS(x1) = x1
MARK(x1) = x1
terms(x1) = terms(x1)
mark(x1) = x1
sqr(x1) = x1
add(x1, x2) = add(x1, x2)
A__ADD(x1, x2) = x2
0 = 0
dbl(x1) = x1
first(x1, x2) = first(x1, x2)
A__FIRST(x1, x2) = A__FIRST(x1, x2)
s(x1) = s
cons(x1, x2) = cons(x1)
recip(x1) = x1
a__terms(x1) = a__terms(x1)
a__sqr(x1) = x1
a__dbl(x1) = x1
a__add(x1, x2) = a__add(x1, x2)
a__first(x1, x2) = a__first(x1, x2)
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

[terms₁, a_{_terms₁}] > cons₁
[add₂, a_{_add₂}] > s > cons₁
0 > cons₁
[first₂, A_{_FIRST₂}, a_{_first₂}, nil] > cons₁

Status:

a_{_first₂}: [2,1]
add₂: [2,1]
cons₁: [1]
a_{_terms₁}: [1]
a_{_add₂}: [2,1]
A_{_FIRST₂}: [2,1]
s: []
0: []
first₂: [2,1]
terms₁: [1]
nil: []

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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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)

(6) Obligation:

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

A__TERMS(N) → MARK(N)
MARK(sqr(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(8) Obligation:

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

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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(sqr(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)
dbl(x1) = x1
sqr(x1) = sqr(x1)
recip(x1) = x1
a__terms(x1) = x1
cons(x1, x2) = cons
a__sqr(x1) = a__sqr(x1)
mark(x1) = x1
terms(x1) = x1
s(x1) = s
0 = 0
add(x1, x2) = x2
a__dbl(x1) = x1
a__add(x1, x2) = x2
a__first(x1, x2) = a__first(x1, x2)
nil = nil
first(x1, x2) = first(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

MARK₁ > [cons, s]
[sqr₁, a_{_sqr₁}] > [cons, s]
[0, a_{_first₂}, nil, first₂] > [cons, s]

Status:

a_{_first₂}: [2,1]
MARK₁: [1]
cons: []
sqr₁: [1]
s: []
0: []
first₂: [2,1]
nil: []
a_{_sqr₁}: [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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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:
The TRS P consists of the following rules:

MARK(dbl(X)) → MARK(X)
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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(dbl(X)) → MARK(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[mark₁, add₁, a_{_add₁}] > [a_{_terms}, a_{_sqr}, s, 0] > [dbl₁, a_{_dbl₁}] > nil
[mark₁, add₁, a_{_add₁}] > [a_{_terms}, a_{_sqr}, s, 0] > terms > nil
[mark₁, add₁, a_{_add₁}] > [a_{_terms}, a_{_sqr}, s, 0] > sqr > nil

Status:

add₁: [1]
a_{_terms}: []
a_{_dbl₁}: [1]
mark₁: [1]
s: []
0: []
a_{_sqr}: []
sqr: []
dbl₁: [1]
a_{_add₁}: [1]
nil: []
terms: []

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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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)

(12) Obligation:

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

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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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.

(13) 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)
a__terms(x1) = a__terms(x1)
cons(x1, x2) = cons(x2)
a__sqr(x1) = x1
mark(x1) = mark(x1)
terms(x1) = terms(x1)
s(x1) = s
0 = 0
add(x1, x2) = add(x1, x2)
sqr(x1) = x1
dbl(x1) = x1
a__dbl(x1) = x1
a__add(x1, x2) = a__add(x1, x2)
a__first(x1, x2) = a__first(x2)
nil = nil
first(x1, x2) = first(x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

MARK₁ > recip₁
[mark₁, add₂, a_{_add₂}, a_{_first₁}, first₁] > a_{_terms₁} > [cons₁, s] > recip₁
[mark₁, add₂, a_{_add₂}, a_{_first₁}, first₁] > a_{_terms₁} > terms₁ > recip₁
[mark₁, add₂, a_{_add₂}, a_{_first₁}, first₁] > 0 > nil > recip₁

Status:

a_{_first₁}: [1]
cons₁: [1]
a_{_add₂}: [2,1]
mark₁: [1]
s: []
0: []
terms₁: [1]
add₂: [2,1]
MARK₁: [1]
a_{_terms₁}: [1]
first₁: [1]
nil: []
recip₁: [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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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)

(14) 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(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, 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(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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) PisEmptyProof (EQUIVALENT transformation)

(16) TRUE