(0) Obligation:

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
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:

G(s(X)) → G(X)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
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 3 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(s(X)) → G(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


G(s(X)) → G(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
0 > s1 > G1 > [cons2, ng1]
sel2 > activate1 > f1 > nf1 > [cons2, ng1]
sel2 > activate1 > g1 > s1 > G1 > [cons2, ng1]

Status:
G1: [1]
s1: [1]
f1: [1]
cons2: [1,2]
nf1: [1]
ng1: [1]
g1: [1]
0: []
sel2: [1,2]
activate1: [1]


The following usable rules [FROCOS05] were oriented:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

(7) Obligation:

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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.


ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
ACTIVATE1 > [ng1, cons2, s1]
sel2 > activate1 > f1 > nf1 > [ng1, cons2, s1]
sel2 > activate1 > g1 > 0 > [ng1, cons2, s1]

Status:
ACTIVATE1: [1]
ng1: [1]
nf1: [1]
f1: [1]
cons2: [1,2]
g1: [1]
0: []
s1: [1]
sel2: [1,2]
activate1: [1]


The following usable rules [FROCOS05] were oriented:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

(12) Obligation:

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
SEL2 > activate1 > f1 > nf1 > [s1, cons2, ng1, 0]
SEL2 > activate1 > g1 > [s1, cons2, ng1, 0]
sel2 > activate1 > f1 > nf1 > [s1, cons2, ng1, 0]
sel2 > activate1 > g1 > [s1, cons2, ng1, 0]

Status:
SEL2: [1,2]
s1: [1]
cons2: [1,2]
activate1: [1]
f1: [1]
nf1: [1]
ng1: [1]
g1: [1]
0: []
sel2: [1,2]


The following usable rules [FROCOS05] were oriented:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

(17) Obligation:

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE