(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(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(X)) → MARK(if(X, c, f(true)))
ACTIVE(f(X)) → IF(X, c, f(true))
ACTIVE(f(X)) → F(true)
ACTIVE(if(true, X, Y)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
MARK(if(X1, X2, X3)) → IF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(c) → ACTIVE(c)
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
F(mark(X)) → F(X)
F(active(X)) → F(X)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
The TRS R consists of the following rules:
active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(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 3 SCCs with 7 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
The TRS R consists of the following rules:
active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(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:
F(active(X)) → F(X)
F(mark(X)) → F(X)
The TRS R consists of the following rules:
active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
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:
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(if(X, c, f(true)))
MARK(f(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(if(X1, X2, X3)) → MARK(X2)
ACTIVE(if(false, X, Y)) → MARK(Y)
The TRS R consists of the following rules:
active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.