Termination w.r.t. Q proof of /tmp/tmp2GVsqC/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 DependencyGraphProof (⇔)

↳12 AND

    ↳13 QDP

        ↳14 QDPOrderProof (⇔)

        ↳15 QDP

        ↳16 DependencyGraphProof (⇔)

        ↳17 TRUE

    ↳18 QDP

        ↳19 QDPOrderProof (⇔)

        ↳20 QDP

        ↳21 PisEmptyProof (⇔)

        ↳22 TRUE

(0) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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:

TERMS(N) → SQR(N)
TERMS(N) → S(N)
SQR(s(X)) → S(n__add(sqr(activate(X)), dbl(activate(X))))
SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → S(n__s(n__dbl(activate(X))))
DBL(s(X)) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__terms(X)) → TERMS(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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 5 less nodes.

(4) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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.

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1) = SQR(x1)
s(x1) = x1
activate(x1) = x1
ACTIVATE(x1) = ACTIVATE(x1)
n__terms(x1) = x1
TERMS(x1) = TERMS(x1)
DBL(x1) = DBL(x1)
n__add(x1, x2) = n__add(x1, x2)
ADD(x1, x2) = ADD(x1)
n__dbl(x1) = x1
n__first(x1, x2) = n__first(x1, x2)
FIRST(x1, x2) = FIRST(x1, x2)
cons(x1, x2) = x2
terms(x1) = x1
add(x1, x2) = add(x1, x2)
n__s(x1) = x1
dbl(x1) = x1
first(x1, x2) = first(x1, x2)
recip(x1) = recip
sqr(x1) = x1
0 = 0
nil = nil

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

[n_{_add₂}, add₂] > [SQR₁, ACTIVATE₁, TERMS₁, DBL₁, ADD₁]
[n_{_first₂}, first₂] > FIRST₂ > [SQR₁, ACTIVATE₁, TERMS₁, DBL₁, ADD₁]
[n_{_first₂}, first₂] > nil
0 > nil

Status:

SQR₁: [1]
ACTIVATE₁: [1]
TERMS₁: [1]
DBL₁: [1]
n_{_add₂}: [1,2]
ADD₁: [1]
n_{_first₂}: [1,2]
FIRST₂: [2,1]
add₂: [1,2]
first₂: [1,2]
recip: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)

(6) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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 1 less node.

(8) Obligation:

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

SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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.

SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
SQR(s(X)) → DBL(activate(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1) = SQR(x1)
s(x1) = x1
ACTIVATE(x1) = ACTIVATE(x1)
n__terms(x1) = n__terms(x1)
TERMS(x1) = TERMS(x1)
activate(x1) = x1
DBL(x1) = DBL(x1)
n__dbl(x1) = x1
terms(x1) = terms(x1)
n__add(x1, x2) = x2
add(x1, x2) = x2
n__s(x1) = x1
dbl(x1) = x1
n__first(x1, x2) = n__first(x1, x2)
first(x1, x2) = first(x1, x2)
cons(x1, x2) = x1
recip(x1) = recip
sqr(x1) = sqr
0 = 0
nil = nil

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

[n_{_terms₁}, terms₁] > [SQR₁, TERMS₁] > [ACTIVATE₁, DBL₁]
[n_{_terms₁}, terms₁] > recip
[n_{_first₂}, first₂]
[sqr, 0, nil]

Status:

SQR₁: [1]
ACTIVATE₁: [1]
n_{_terms₁}: [1]
TERMS₁: [1]
DBL₁: [1]
terms₁: [1]
n_{_first₂}: [2,1]
first₂: [2,1]
recip: []
sqr: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)

(10) Obligation:

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

TERMS(N) → SQR(N)
SQR(s(X)) → SQR(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(12) Complex Obligation (AND)

(13) Obligation:

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

ACTIVATE(n__dbl(X)) → DBL(X)
DBL(s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = ACTIVATE(x1)
n__dbl(x1) = x1
DBL(x1) = DBL(x1)
s(x1) = s(x1)

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

[ACTIVATE₁, DBL₁, s₁]

Status:

ACTIVATE₁: [1]
DBL₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

ACTIVATE(n__dbl(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(16) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(17) TRUE

(18) Obligation:

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

SQR(s(X)) → SQR(activate(X))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SQR(s(X)) → SQR(activate(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1) = x1
s(x1) = s(x1)
activate(x1) = activate(x1)
n__terms(x1) = n__terms(x1)
terms(x1) = terms(x1)
n__add(x1, x2) = n__add(x1, x2)
add(x1, x2) = add(x1, x2)
n__s(x1) = n__s(x1)
n__dbl(x1) = n__dbl(x1)
dbl(x1) = dbl(x1)
n__first(x1, x2) = x2
first(x1, x2) = first(x2)
cons(x1, x2) = cons
recip(x1) = recip
sqr(x1) = sqr
0 = 0
nil = nil

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

[n_{_add₂}, add₂] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > n_{_terms₁}
[n_{_add₂}, add₂] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > cons
[n_{_add₂}, add₂] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > recip
[n_{_add₂}, add₂] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > 0
[n_{_add₂}, add₂] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > nil
[n_{_dbl₁}, dbl₁] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > n_{_terms₁}
[n_{_dbl₁}, dbl₁] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > cons
[n_{_dbl₁}, dbl₁] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > terms₁ > recip
[n_{_dbl₁}, dbl₁] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > 0
[n_{_dbl₁}, dbl₁] > [s₁, n_{_s₁}] > [activate₁, first₁, sqr] > nil

Status:

s₁: [1]
activate₁: [1]
n_{_terms₁}: [1]
terms₁: [1]
n_{_add₂}: [2,1]
add₂: [2,1]
n_{_s₁}: [1]
n_{_dbl₁}: [1]
dbl₁: [1]
first₁: [1]
cons: []
recip: []
sqr: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)

(20) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(21) 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) DependencyGraphProof (EQUIVALENT transformation)

(12) Complex Obligation (AND)

(13) Obligation:

(14) QDPOrderProof (EQUIVALENT transformation)

(15) Obligation:

(16) DependencyGraphProof (EQUIVALENT transformation)

(17) TRUE

(18) Obligation:

(19) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) TRUE