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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

g(cons(s(X), Y)) → s(X)
g(cons(0, Y)) → g(Y)

The TRS R 2 is

f(s(X)) → f(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The signature Sigma is {f, h}

(2) Obligation:

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

F(s(X)) → F(X)
G(cons(0, Y)) → G(Y)
H(cons(X, Y)) → H(g(cons(X, Y)))
H(cons(X, Y)) → G(cons(X, Y))

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

G(cons(0, Y)) → G(Y)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(cons(0, Y)) → G(Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  G(x1)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
f(x1)  =  f
s(x1)  =  s(x1)
g(x1)  =  x1
h(x1)  =  h

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

0 > G₁ > cons₂
f > cons₂
s₁ > cons₂
h > cons₂

Status:

cons₂: [2,1]
f: []
G₁: [1]
s₁: [1]
h: []
0: []

The following usable rules [FROCOS05] were oriented:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

(9) Obligation:

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(s(X)) → F(X)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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.

F(s(X)) → F(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
s(x1)  =  s(x1)
f(x1)  =  f
g(x1)  =  g(x1)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
h(x1)  =  h

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

F₁ > [f, cons₂]
s₁ > [f, cons₂]
h > [g₁, 0] > [f, cons₂]

Status:

cons₂: [2,1]
f: []
g₁: [1]
s₁: [1]
h: []
0: []
F₁: [1]

The following usable rules [FROCOS05] were oriented:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

(14) Obligation:

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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.

(16) TRUE