(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.