(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)
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(f(a, b, X)) → MARK(f(X, X, X))
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(c) → MARK(a)
ACTIVE(c) → MARK(b)
MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3)))
MARK(f(X1, X2, X3)) → F(mark(X1), X2, mark(X3))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → MARK(X3)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
MARK(c) → ACTIVE(c)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
The TRS R consists of the following rules:
active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)
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 7 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
The TRS R consists of the following rules:
active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)
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:
MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → MARK(X3)
The TRS R consists of the following rules:
active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.