Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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:

H(e(x), y) → H(d(x, y), s(y))
H(e(x), y) → D(x, y)
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
D(g(x, y), z) → G(d(x, z), e(y))
D(g(x, y), z) → D(x, z)
G(e(x), e(y)) → G(x, y)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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(e(x), e(y)) → G(x, y)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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(e(x), e(y)) → G(x, y)

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

Recursive Path Order [RPO].
Precedence:

0 > [G₁, e₁, d₁] > [h₂, s]

The following usable rules [FROCOS05] were oriented:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

(7) Obligation:

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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:

D(g(x, y), z) → D(x, z)
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

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

Recursive Path Order [RPO].
Precedence:

0 > [D₁, s₁]
h₁ > [D₁, s₁]

The following usable rules [FROCOS05] were oriented:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

(12) Obligation:

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

D(g(x, y), z) → D(x, z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.

D(g(x, y), z) → D(x, z)

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

Recursive Path Order [RPO].
Precedence:

h > d₂ > [D₂, g₂] > [e, s₁] > 0

The following usable rules [FROCOS05] were oriented:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

(14) Obligation:

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.

(16) TRUE

(17) Obligation:

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

H(e(x), y) → H(d(x, y), s(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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