(0) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → 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(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))
G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → 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 2 SCCs.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → 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(X, s(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(x2)
s(x1)  =  s(x1)
h(x1, x2)  =  h(x1, x2)
f(x1, x2, x3)  =  f(x1, x3)
g(x1, x2)  =  g
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > [G1, h2, f2, g, 0]

Status:
G1: [1]
s1: [1]
h2: [2,1]
f2: [2,1]
g: []
0: []


The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → 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:

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(X, Z) → F(X, s(X), Z)

The TRS R consists of the following rules:

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

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