3.81/1.85	YES
3.96/1.86	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.96/1.86	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.96/1.86	
3.96/1.86	
3.96/1.86	Termination w.r.t. Q of the given QTRS could be proven:
3.96/1.86	
3.96/1.86	(0) QTRS
3.96/1.86	(1) DependencyPairsProof [EQUIVALENT, 2 ms]
3.96/1.86	(2) QDP
3.96/1.86	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.96/1.86	(4) AND
3.96/1.86	    (5) QDP
3.96/1.86	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.96/1.86	        (7) QDP
3.96/1.86	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.96/1.86	        (9) QDP
3.96/1.86	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.96/1.86	        (11) YES
3.96/1.86	    (12) QDP
3.96/1.86	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
3.96/1.86	        (14) QDP
3.96/1.86	        (15) QReductionProof [EQUIVALENT, 0 ms]
3.96/1.86	        (16) QDP
3.96/1.86	        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.96/1.86	        (18) YES
3.96/1.86	    (19) QDP
3.96/1.86	        (20) UsableRulesProof [EQUIVALENT, 0 ms]
3.96/1.86	        (21) QDP
3.96/1.86	        (22) QReductionProof [EQUIVALENT, 0 ms]
3.96/1.86	        (23) QDP
3.96/1.86	        (24) TransformationProof [EQUIVALENT, 0 ms]
3.96/1.86	        (25) QDP
3.96/1.86	        (26) DependencyGraphProof [EQUIVALENT, 0 ms]
3.96/1.86	        (27) TRUE
3.96/1.86	    (28) QDP
3.96/1.86	        (29) UsableRulesProof [EQUIVALENT, 0 ms]
3.96/1.86	        (30) QDP
3.96/1.86	        (31) QReductionProof [EQUIVALENT, 0 ms]
3.96/1.86	        (32) QDP
3.96/1.86	        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.96/1.86	        (34) YES
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(0)
3.96/1.86	Obligation:
3.96/1.86	Q restricted rewrite system:
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(1) DependencyPairsProof (EQUIVALENT)
3.96/1.86	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(2)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> MARK(f(Y, Y, Y))
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> F(Y, Y, Y)
3.96/1.86	   ACTIVE(g(b)) -> MARK(c)
3.96/1.86	   ACTIVE(b) -> MARK(c)
3.96/1.86	   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, X3))
3.96/1.86	   MARK(g(X)) -> ACTIVE(g(mark(X)))
3.96/1.86	   MARK(g(X)) -> G(mark(X))
3.96/1.86	   MARK(g(X)) -> MARK(X)
3.96/1.86	   MARK(b) -> ACTIVE(b)
3.96/1.86	   MARK(c) -> ACTIVE(c)
3.96/1.86	   F(mark(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, mark(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
3.96/1.86	   F(active(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, active(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, active(X3)) -> F(X1, X2, X3)
3.96/1.86	   G(mark(X)) -> G(X)
3.96/1.86	   G(active(X)) -> G(X)
3.96/1.86	
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(3) DependencyGraphProof (EQUIVALENT)
3.96/1.86	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 7 less nodes.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(4)
3.96/1.86	Complex Obligation (AND)
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(5)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   G(active(X)) -> G(X)
3.96/1.86	   G(mark(X)) -> G(X)
3.96/1.86	
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(6) UsableRulesProof (EQUIVALENT)
3.96/1.86	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(7)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   G(active(X)) -> G(X)
3.96/1.86	   G(mark(X)) -> G(X)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(8) QReductionProof (EQUIVALENT)
3.96/1.86	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.96/1.86	
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(9)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   G(active(X)) -> G(X)
3.96/1.86	   G(mark(X)) -> G(X)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(10) QDPSizeChangeProof (EQUIVALENT)
3.96/1.86	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.96/1.86	
3.96/1.86	From the DPs we obtained the following set of size-change graphs:
3.96/1.86	*G(active(X)) -> G(X)
3.96/1.86	The graph contains the following edges 1 > 1
3.96/1.86	
3.96/1.86	
3.96/1.86	*G(mark(X)) -> G(X)
3.96/1.86	The graph contains the following edges 1 > 1
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(11)
3.96/1.86	YES
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(12)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   F(X1, mark(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(mark(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
3.96/1.86	   F(active(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, active(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, active(X3)) -> F(X1, X2, X3)
3.96/1.86	
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(13) UsableRulesProof (EQUIVALENT)
3.96/1.86	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(14)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   F(X1, mark(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(mark(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
3.96/1.86	   F(active(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, active(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, active(X3)) -> F(X1, X2, X3)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(15) QReductionProof (EQUIVALENT)
3.96/1.86	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.96/1.86	
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(16)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   F(X1, mark(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(mark(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
3.96/1.86	   F(active(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, active(X2), X3) -> F(X1, X2, X3)
3.96/1.86	   F(X1, X2, active(X3)) -> F(X1, X2, X3)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(17) QDPSizeChangeProof (EQUIVALENT)
3.96/1.86	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.96/1.86	
3.96/1.86	From the DPs we obtained the following set of size-change graphs:
3.96/1.86	*F(X1, mark(X2), X3) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
3.96/1.86	
3.96/1.86	
3.96/1.86	*F(mark(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
3.96/1.86	
3.96/1.86	
3.96/1.86	*F(X1, X2, mark(X3)) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
3.96/1.86	
3.96/1.86	
3.96/1.86	*F(active(X1), X2, X3) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
3.96/1.86	
3.96/1.86	
3.96/1.86	*F(X1, active(X2), X3) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
3.96/1.86	
3.96/1.86	
3.96/1.86	*F(X1, X2, active(X3)) -> F(X1, X2, X3)
3.96/1.86	The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(18)
3.96/1.86	YES
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(19)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, X3))
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> MARK(f(Y, Y, Y))
3.96/1.86	
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(20) UsableRulesProof (EQUIVALENT)
3.96/1.86	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(21)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, X3))
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> MARK(f(Y, Y, Y))
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(22) QReductionProof (EQUIVALENT)
3.96/1.86	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(23)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, X3))
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> MARK(f(Y, Y, Y))
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(24) TransformationProof (EQUIVALENT)
3.96/1.86	By instantiating [LPAR04] the rule MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, X3)) we obtained the following new rules [LPAR04]:
3.96/1.86	
3.96/1.86	   (MARK(f(z1, z1, z1)) -> ACTIVE(f(z1, z1, z1)),MARK(f(z1, z1, z1)) -> ACTIVE(f(z1, z1, z1)))
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(25)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   ACTIVE(f(X, g(X), Y)) -> MARK(f(Y, Y, Y))
3.96/1.86	   MARK(f(z1, z1, z1)) -> ACTIVE(f(z1, z1, z1))
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(26) DependencyGraphProof (EQUIVALENT)
3.96/1.86	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(27)
3.96/1.86	TRUE
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(28)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(g(X)) -> MARK(X)
3.96/1.86	
3.96/1.86	The TRS R consists of the following rules:
3.96/1.86	
3.96/1.86	   active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
3.96/1.86	   active(g(b)) -> mark(c)
3.96/1.86	   active(b) -> mark(c)
3.96/1.86	   mark(f(X1, X2, X3)) -> active(f(X1, X2, X3))
3.96/1.86	   mark(g(X)) -> active(g(mark(X)))
3.96/1.86	   mark(b) -> active(b)
3.96/1.86	   mark(c) -> active(c)
3.96/1.86	   f(mark(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, mark(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
3.96/1.86	   f(active(X1), X2, X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, active(X2), X3) -> f(X1, X2, X3)
3.96/1.86	   f(X1, X2, active(X3)) -> f(X1, X2, X3)
3.96/1.86	   g(mark(X)) -> g(X)
3.96/1.86	   g(active(X)) -> g(X)
3.96/1.86	
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(29) UsableRulesProof (EQUIVALENT)
3.96/1.86	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(30)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(g(X)) -> MARK(X)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(31) QReductionProof (EQUIVALENT)
3.96/1.86	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.96/1.86	
3.96/1.86	   active(f(x0, g(x0), x1))
3.96/1.86	   active(g(b))
3.96/1.86	   active(b)
3.96/1.86	   mark(f(x0, x1, x2))
3.96/1.86	   mark(g(x0))
3.96/1.86	   mark(b)
3.96/1.86	   mark(c)
3.96/1.86	   f(mark(x0), x1, x2)
3.96/1.86	   f(x0, mark(x1), x2)
3.96/1.86	   f(x0, x1, mark(x2))
3.96/1.86	   f(active(x0), x1, x2)
3.96/1.86	   f(x0, active(x1), x2)
3.96/1.86	   f(x0, x1, active(x2))
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(32)
3.96/1.86	Obligation:
3.96/1.86	Q DP problem:
3.96/1.86	The TRS P consists of the following rules:
3.96/1.86	
3.96/1.86	   MARK(g(X)) -> MARK(X)
3.96/1.86	
3.96/1.86	R is empty.
3.96/1.86	The set Q consists of the following terms:
3.96/1.86	
3.96/1.86	   g(mark(x0))
3.96/1.86	   g(active(x0))
3.96/1.86	
3.96/1.86	We have to consider all minimal (P,Q,R)-chains.
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(33) QDPSizeChangeProof (EQUIVALENT)
3.96/1.86	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.96/1.86	
3.96/1.86	From the DPs we obtained the following set of size-change graphs:
3.96/1.86	*MARK(g(X)) -> MARK(X)
3.96/1.86	The graph contains the following edges 1 > 1
3.96/1.86	
3.96/1.86	
3.96/1.86	----------------------------------------
3.96/1.86	
3.96/1.86	(34)
3.96/1.86	YES
3.96/1.89	EOF