(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(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:
F(j(x, y), y) → G(f(x, k(y)))
F(j(x, y), y) → F(x, k(y))
F(j(x, y), y) → K(y)
F(x, h1(y, z)) → H2(0, x, h1(y, z))
G(h2(x, y, h1(z, u))) → H2(s(x), y, h1(z, u))
H2(x, j(y, h1(z, u)), h1(z, u)) → H2(s(x), y, h1(s(z), u))
The TRS R consists of the following rules:
f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(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 with 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H2(x, j(y, h1(z, u)), h1(z, u)) → H2(s(x), y, h1(s(z), u))
The TRS R consists of the following rules:
f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(j(x, y), y) → F(x, k(y))
The TRS R consists of the following rules:
f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.