(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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:
ACTIVE(h(X)) → MARK(g(X, X))
ACTIVE(h(X)) → G(X, X)
ACTIVE(g(a, X)) → MARK(f(b, X))
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(f(X, X)) → MARK(h(a))
ACTIVE(f(X, X)) → H(a)
ACTIVE(a) → MARK(b)
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → H(mark(X))
MARK(h(X)) → MARK(X)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)
MARK(a) → ACTIVE(a)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(b) → ACTIVE(b)
H(mark(X)) → H(X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
G(X1, mark(X2)) → G(X1, X2)
G(active(X1), X2) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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 4 SCCs with 9 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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:
G(X1, mark(X2)) → G(X1, X2)
G(mark(X1), X2) → G(X1, X2)
G(active(X1), X2) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(active(X)) → H(X)
H(mark(X)) → H(X)
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(h(X)) → ACTIVE(h(mark(X)))
ACTIVE(h(X)) → MARK(g(X, X))
MARK(h(X)) → MARK(X)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
ACTIVE(g(a, X)) → MARK(f(b, X))
MARK(g(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(h(a))
MARK(f(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.