6.72/2.64	YES
6.72/2.65	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
6.72/2.65	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
6.72/2.65	
6.72/2.65	
6.72/2.65	Termination w.r.t. Q of the given QTRS could be proven:
6.72/2.65	
6.72/2.65	(0) QTRS
6.72/2.65	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
6.72/2.65	(2) QDP
6.72/2.65	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
6.72/2.65	(4) AND
6.72/2.65	    (5) QDP
6.72/2.65	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
6.72/2.65	        (7) QDP
6.72/2.65	        (8) QReductionProof [EQUIVALENT, 0 ms]
6.72/2.65	        (9) QDP
6.72/2.65	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
6.72/2.65	        (11) YES
6.72/2.65	    (12) QDP
6.72/2.65	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
6.72/2.65	        (14) QDP
6.72/2.65	        (15) QReductionProof [EQUIVALENT, 0 ms]
6.72/2.65	        (16) QDP
6.72/2.65	        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
6.72/2.65	        (18) YES
6.72/2.65	    (19) QDP
6.72/2.65	        (20) UsableRulesProof [EQUIVALENT, 0 ms]
6.72/2.65	        (21) QDP
6.72/2.65	        (22) QReductionProof [EQUIVALENT, 0 ms]
6.72/2.65	        (23) QDP
6.72/2.65	        (24) QDPSizeChangeProof [EQUIVALENT, 0 ms]
6.72/2.65	        (25) YES
6.72/2.65	    (26) QDP
6.72/2.65	        (27) UsableRulesProof [EQUIVALENT, 0 ms]
6.72/2.65	        (28) QDP
6.72/2.65	        (29) QReductionProof [EQUIVALENT, 0 ms]
6.72/2.65	        (30) QDP
6.72/2.65	        (31) QDPSizeChangeProof [EQUIVALENT, 0 ms]
6.72/2.65	        (32) YES
6.72/2.65	    (33) QDP
6.72/2.65	        (34) MRRProof [EQUIVALENT, 54 ms]
6.72/2.65	        (35) QDP
6.72/2.65	        (36) QDPOrderProof [EQUIVALENT, 94 ms]
6.72/2.65	        (37) QDP
6.72/2.65	        (38) DependencyGraphProof [EQUIVALENT, 0 ms]
6.72/2.65	        (39) AND
6.72/2.65	            (40) QDP
6.72/2.65	                (41) TransformationProof [EQUIVALENT, 0 ms]
6.72/2.65	                (42) QDP
6.72/2.65	                (43) QDPOrderProof [EQUIVALENT, 0 ms]
6.72/2.65	                (44) QDP
6.72/2.65	                (45) PisEmptyProof [EQUIVALENT, 0 ms]
6.72/2.65	                (46) YES
6.72/2.65	            (47) QDP
6.72/2.65	                (48) UsableRulesProof [EQUIVALENT, 0 ms]
6.72/2.65	                (49) QDP
6.72/2.65	                (50) QReductionProof [EQUIVALENT, 0 ms]
6.72/2.65	                (51) QDP
6.72/2.65	                (52) QDPSizeChangeProof [EQUIVALENT, 0 ms]
6.72/2.65	                (53) YES
6.72/2.65	
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(0)
6.72/2.65	Obligation:
6.72/2.65	Q restricted rewrite system:
6.72/2.65	The TRS R consists of the following rules:
6.72/2.65	
6.72/2.65	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.65	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.65	   active(p(s(0))) -> mark(0)
6.72/2.65	   mark(f(X)) -> active(f(mark(X)))
6.72/2.65	   mark(0) -> active(0)
6.72/2.65	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.65	   mark(s(X)) -> active(s(mark(X)))
6.72/2.65	   mark(p(X)) -> active(p(mark(X)))
6.72/2.65	   f(mark(X)) -> f(X)
6.72/2.65	   f(active(X)) -> f(X)
6.72/2.65	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.65	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.65	   s(mark(X)) -> s(X)
6.72/2.65	   s(active(X)) -> s(X)
6.72/2.65	   p(mark(X)) -> p(X)
6.72/2.65	   p(active(X)) -> p(X)
6.72/2.65	
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(1) DependencyPairsProof (EQUIVALENT)
6.72/2.65	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(2)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   ACTIVE(f(0)) -> MARK(cons(0, f(s(0))))
6.72/2.65	   ACTIVE(f(0)) -> CONS(0, f(s(0)))
6.72/2.65	   ACTIVE(f(0)) -> F(s(0))
6.72/2.65	   ACTIVE(f(0)) -> S(0)
6.72/2.65	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.65	   ACTIVE(f(s(0))) -> F(p(s(0)))
6.72/2.65	   ACTIVE(f(s(0))) -> P(s(0))
6.72/2.65	   ACTIVE(p(s(0))) -> MARK(0)
6.72/2.65	   MARK(f(X)) -> ACTIVE(f(mark(X)))
6.72/2.65	   MARK(f(X)) -> F(mark(X))
6.72/2.65	   MARK(f(X)) -> MARK(X)
6.72/2.65	   MARK(0) -> ACTIVE(0)
6.72/2.65	   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
6.72/2.65	   MARK(cons(X1, X2)) -> CONS(mark(X1), X2)
6.72/2.65	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.65	   MARK(s(X)) -> ACTIVE(s(mark(X)))
6.72/2.65	   MARK(s(X)) -> S(mark(X))
6.72/2.65	   MARK(s(X)) -> MARK(X)
6.72/2.65	   MARK(p(X)) -> ACTIVE(p(mark(X)))
6.72/2.65	   MARK(p(X)) -> P(mark(X))
6.72/2.65	   MARK(p(X)) -> MARK(X)
6.72/2.65	   F(mark(X)) -> F(X)
6.72/2.65	   F(active(X)) -> F(X)
6.72/2.65	   CONS(mark(X1), X2) -> CONS(X1, X2)
6.72/2.65	   CONS(X1, mark(X2)) -> CONS(X1, X2)
6.72/2.65	   CONS(active(X1), X2) -> CONS(X1, X2)
6.72/2.65	   CONS(X1, active(X2)) -> CONS(X1, X2)
6.72/2.65	   S(mark(X)) -> S(X)
6.72/2.65	   S(active(X)) -> S(X)
6.72/2.65	   P(mark(X)) -> P(X)
6.72/2.65	   P(active(X)) -> P(X)
6.72/2.65	
6.72/2.65	The TRS R consists of the following rules:
6.72/2.65	
6.72/2.65	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.65	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.65	   active(p(s(0))) -> mark(0)
6.72/2.65	   mark(f(X)) -> active(f(mark(X)))
6.72/2.65	   mark(0) -> active(0)
6.72/2.65	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.65	   mark(s(X)) -> active(s(mark(X)))
6.72/2.65	   mark(p(X)) -> active(p(mark(X)))
6.72/2.65	   f(mark(X)) -> f(X)
6.72/2.65	   f(active(X)) -> f(X)
6.72/2.65	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.65	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.65	   s(mark(X)) -> s(X)
6.72/2.65	   s(active(X)) -> s(X)
6.72/2.65	   p(mark(X)) -> p(X)
6.72/2.65	   p(active(X)) -> p(X)
6.72/2.65	
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(3) DependencyGraphProof (EQUIVALENT)
6.72/2.65	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 11 less nodes.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(4)
6.72/2.65	Complex Obligation (AND)
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(5)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   P(active(X)) -> P(X)
6.72/2.65	   P(mark(X)) -> P(X)
6.72/2.65	
6.72/2.65	The TRS R consists of the following rules:
6.72/2.65	
6.72/2.65	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.65	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.65	   active(p(s(0))) -> mark(0)
6.72/2.65	   mark(f(X)) -> active(f(mark(X)))
6.72/2.65	   mark(0) -> active(0)
6.72/2.65	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.65	   mark(s(X)) -> active(s(mark(X)))
6.72/2.65	   mark(p(X)) -> active(p(mark(X)))
6.72/2.65	   f(mark(X)) -> f(X)
6.72/2.65	   f(active(X)) -> f(X)
6.72/2.65	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.65	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.65	   s(mark(X)) -> s(X)
6.72/2.65	   s(active(X)) -> s(X)
6.72/2.65	   p(mark(X)) -> p(X)
6.72/2.65	   p(active(X)) -> p(X)
6.72/2.65	
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(6) UsableRulesProof (EQUIVALENT)
6.72/2.65	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.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(7)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   P(active(X)) -> P(X)
6.72/2.65	   P(mark(X)) -> P(X)
6.72/2.65	
6.72/2.65	R is empty.
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(8) QReductionProof (EQUIVALENT)
6.72/2.65	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.72/2.65	
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(9)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   P(active(X)) -> P(X)
6.72/2.65	   P(mark(X)) -> P(X)
6.72/2.65	
6.72/2.65	R is empty.
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(10) QDPSizeChangeProof (EQUIVALENT)
6.72/2.65	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. 
6.72/2.65	
6.72/2.65	From the DPs we obtained the following set of size-change graphs:
6.72/2.65	*P(active(X)) -> P(X)
6.72/2.65	The graph contains the following edges 1 > 1
6.72/2.65	
6.72/2.65	
6.72/2.65	*P(mark(X)) -> P(X)
6.72/2.65	The graph contains the following edges 1 > 1
6.72/2.65	
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(11)
6.72/2.65	YES
6.72/2.65	
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(12)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   S(active(X)) -> S(X)
6.72/2.65	   S(mark(X)) -> S(X)
6.72/2.65	
6.72/2.65	The TRS R consists of the following rules:
6.72/2.65	
6.72/2.65	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.65	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.65	   active(p(s(0))) -> mark(0)
6.72/2.65	   mark(f(X)) -> active(f(mark(X)))
6.72/2.65	   mark(0) -> active(0)
6.72/2.65	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.65	   mark(s(X)) -> active(s(mark(X)))
6.72/2.65	   mark(p(X)) -> active(p(mark(X)))
6.72/2.65	   f(mark(X)) -> f(X)
6.72/2.65	   f(active(X)) -> f(X)
6.72/2.65	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.65	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.65	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.65	   s(mark(X)) -> s(X)
6.72/2.65	   s(active(X)) -> s(X)
6.72/2.65	   p(mark(X)) -> p(X)
6.72/2.65	   p(active(X)) -> p(X)
6.72/2.65	
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(13) UsableRulesProof (EQUIVALENT)
6.72/2.65	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.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(14)
6.72/2.65	Obligation:
6.72/2.65	Q DP problem:
6.72/2.65	The TRS P consists of the following rules:
6.72/2.65	
6.72/2.65	   S(active(X)) -> S(X)
6.72/2.65	   S(mark(X)) -> S(X)
6.72/2.65	
6.72/2.65	R is empty.
6.72/2.65	The set Q consists of the following terms:
6.72/2.65	
6.72/2.65	   active(f(0))
6.72/2.65	   active(f(s(0)))
6.72/2.65	   active(p(s(0)))
6.72/2.65	   mark(f(x0))
6.72/2.65	   mark(0)
6.72/2.65	   mark(cons(x0, x1))
6.72/2.65	   mark(s(x0))
6.72/2.65	   mark(p(x0))
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.65	   s(active(x0))
6.72/2.65	   p(mark(x0))
6.72/2.65	   p(active(x0))
6.72/2.65	
6.72/2.65	We have to consider all minimal (P,Q,R)-chains.
6.72/2.65	----------------------------------------
6.72/2.65	
6.72/2.65	(15) QReductionProof (EQUIVALENT)
6.72/2.65	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.72/2.65	
6.72/2.65	   f(mark(x0))
6.72/2.65	   f(active(x0))
6.72/2.65	   cons(mark(x0), x1)
6.72/2.65	   cons(x0, mark(x1))
6.72/2.65	   cons(active(x0), x1)
6.72/2.65	   cons(x0, active(x1))
6.72/2.65	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(16)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   S(active(X)) -> S(X)
6.72/2.66	   S(mark(X)) -> S(X)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(17) QDPSizeChangeProof (EQUIVALENT)
6.72/2.66	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. 
6.72/2.66	
6.72/2.66	From the DPs we obtained the following set of size-change graphs:
6.72/2.66	*S(active(X)) -> S(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	*S(mark(X)) -> S(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(18)
6.72/2.66	YES
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(19)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   CONS(X1, mark(X2)) -> CONS(X1, X2)
6.72/2.66	   CONS(mark(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(active(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(X1, active(X2)) -> CONS(X1, X2)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(20) UsableRulesProof (EQUIVALENT)
6.72/2.66	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.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(21)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   CONS(X1, mark(X2)) -> CONS(X1, X2)
6.72/2.66	   CONS(mark(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(active(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(X1, active(X2)) -> CONS(X1, X2)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(22) QReductionProof (EQUIVALENT)
6.72/2.66	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.72/2.66	
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(23)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   CONS(X1, mark(X2)) -> CONS(X1, X2)
6.72/2.66	   CONS(mark(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(active(X1), X2) -> CONS(X1, X2)
6.72/2.66	   CONS(X1, active(X2)) -> CONS(X1, X2)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(24) QDPSizeChangeProof (EQUIVALENT)
6.72/2.66	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. 
6.72/2.66	
6.72/2.66	From the DPs we obtained the following set of size-change graphs:
6.72/2.66	*CONS(X1, mark(X2)) -> CONS(X1, X2)
6.72/2.66	The graph contains the following edges 1 >= 1, 2 > 2
6.72/2.66	
6.72/2.66	
6.72/2.66	*CONS(mark(X1), X2) -> CONS(X1, X2)
6.72/2.66	The graph contains the following edges 1 > 1, 2 >= 2
6.72/2.66	
6.72/2.66	
6.72/2.66	*CONS(active(X1), X2) -> CONS(X1, X2)
6.72/2.66	The graph contains the following edges 1 > 1, 2 >= 2
6.72/2.66	
6.72/2.66	
6.72/2.66	*CONS(X1, active(X2)) -> CONS(X1, X2)
6.72/2.66	The graph contains the following edges 1 >= 1, 2 > 2
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(25)
6.72/2.66	YES
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(26)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   F(active(X)) -> F(X)
6.72/2.66	   F(mark(X)) -> F(X)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(27) UsableRulesProof (EQUIVALENT)
6.72/2.66	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.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(28)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   F(active(X)) -> F(X)
6.72/2.66	   F(mark(X)) -> F(X)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(29) QReductionProof (EQUIVALENT)
6.72/2.66	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.72/2.66	
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(30)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   F(active(X)) -> F(X)
6.72/2.66	   F(mark(X)) -> F(X)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(31) QDPSizeChangeProof (EQUIVALENT)
6.72/2.66	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. 
6.72/2.66	
6.72/2.66	From the DPs we obtained the following set of size-change graphs:
6.72/2.66	*F(active(X)) -> F(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	*F(mark(X)) -> F(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(32)
6.72/2.66	YES
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(33)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
6.72/2.66	   ACTIVE(f(0)) -> MARK(cons(0, f(s(0))))
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(f(X)) -> ACTIVE(f(mark(X)))
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(f(X)) -> MARK(X)
6.72/2.66	   MARK(s(X)) -> ACTIVE(s(mark(X)))
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(p(X)) -> ACTIVE(p(mark(X)))
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(34) MRRProof (EQUIVALENT)
6.72/2.66	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
6.72/2.66	
6.72/2.66	Strictly oriented dependency pairs:
6.72/2.66	
6.72/2.66	   MARK(f(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	
6.72/2.66	Used ordering: Polynomial interpretation [POLO]:
6.72/2.66	
6.72/2.66	   POL(0) = 0
6.72/2.66	   POL(ACTIVE(x_1)) = x_1
6.72/2.66	   POL(MARK(x_1)) = x_1
6.72/2.66	   POL(active(x_1)) = x_1
6.72/2.66	   POL(cons(x_1, x_2)) = x_1 + x_2
6.72/2.66	   POL(f(x_1)) = 2 + 2*x_1
6.72/2.66	   POL(mark(x_1)) = x_1
6.72/2.66	   POL(p(x_1)) = x_1
6.72/2.66	   POL(s(x_1)) = x_1
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(35)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
6.72/2.66	   ACTIVE(f(0)) -> MARK(cons(0, f(s(0))))
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(f(X)) -> ACTIVE(f(mark(X)))
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(s(X)) -> ACTIVE(s(mark(X)))
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(p(X)) -> ACTIVE(p(mark(X)))
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(36) QDPOrderProof (EQUIVALENT)
6.72/2.66	We use the reduction pair processor [LPAR04,JAR06].
6.72/2.66	
6.72/2.66	
6.72/2.66	The following pairs can be oriented strictly and are deleted.
6.72/2.66	
6.72/2.66	   ACTIVE(f(0)) -> MARK(cons(0, f(s(0))))
6.72/2.66	The remaining pairs can at least be oriented weakly.
6.72/2.66	Used ordering:  Combined order from the following AFS and order.
6.72/2.66	MARK(x1)  =  x1
6.72/2.66	
6.72/2.66	cons(x1, x2)  =  x1
6.72/2.66	
6.72/2.66	ACTIVE(x1)  =  x1
6.72/2.66	
6.72/2.66	mark(x1)  =  x1
6.72/2.66	
6.72/2.66	f(x1)  =  f
6.72/2.66	
6.72/2.66	0  =  0
6.72/2.66	
6.72/2.66	s(x1)  =  x1
6.72/2.66	
6.72/2.66	p(x1)  =  x1
6.72/2.66	
6.72/2.66	active(x1)  =  x1
6.72/2.66	
6.72/2.66	
6.72/2.66	Knuth-Bendix order [KBO] with precedence:f > 0
6.72/2.66	
6.72/2.66	and weight map:
6.72/2.66	
6.72/2.66	   0=1
6.72/2.66	   f=1
6.72/2.66	
6.72/2.66	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
6.72/2.66	
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(37)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(f(X)) -> ACTIVE(f(mark(X)))
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(s(X)) -> ACTIVE(s(mark(X)))
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(p(X)) -> ACTIVE(p(mark(X)))
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(38) DependencyGraphProof (EQUIVALENT)
6.72/2.66	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(39)
6.72/2.66	Complex Obligation (AND)
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(40)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(f(X)) -> ACTIVE(f(mark(X)))
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(41) TransformationProof (EQUIVALENT)
6.72/2.66	By instantiating [LPAR04] the rule MARK(f(X)) -> ACTIVE(f(mark(X))) we obtained the following new rules [LPAR04]:
6.72/2.66	
6.72/2.66	   (MARK(f(p(s(0)))) -> ACTIVE(f(mark(p(s(0))))),MARK(f(p(s(0)))) -> ACTIVE(f(mark(p(s(0))))))
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(42)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(f(p(s(0)))) -> ACTIVE(f(mark(p(s(0)))))
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(43) QDPOrderProof (EQUIVALENT)
6.72/2.66	We use the reduction pair processor [LPAR04,JAR06].
6.72/2.66	
6.72/2.66	
6.72/2.66	The following pairs can be oriented strictly and are deleted.
6.72/2.66	
6.72/2.66	   ACTIVE(f(s(0))) -> MARK(f(p(s(0))))
6.72/2.66	   MARK(f(p(s(0)))) -> ACTIVE(f(mark(p(s(0)))))
6.72/2.66	The remaining pairs can at least be oriented weakly.
6.72/2.66	Used ordering:  Combined order from the following AFS and order.
6.72/2.66	ACTIVE(x1)  =  x1
6.72/2.66	
6.72/2.66	f(x1)  =  x1
6.72/2.66	
6.72/2.66	s(x1)  =  s
6.72/2.66	
6.72/2.66	MARK(x1)  =  MARK
6.72/2.66	
6.72/2.66	p(x1)  =  p
6.72/2.66	
6.72/2.66	mark(x1)  =  x1
6.72/2.66	
6.72/2.66	active(x1)  =  x1
6.72/2.66	
6.72/2.66	cons(x1, x2)  =  cons
6.72/2.66	
6.72/2.66	0  =  0
6.72/2.66	
6.72/2.66	
6.72/2.66	Knuth-Bendix order [KBO] with precedence:s > MARK > p > 0 > cons
6.72/2.66	
6.72/2.66	and weight map:
6.72/2.66	
6.72/2.66	   s=1
6.72/2.66	   0=1
6.72/2.66	   p=1
6.72/2.66	   cons=1
6.72/2.66	   MARK=1
6.72/2.66	
6.72/2.66	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
6.72/2.66	
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(44)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	P is empty.
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(45) PisEmptyProof (EQUIVALENT)
6.72/2.66	The TRS P is empty. Hence, there is no (P,Q,R) chain.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(46)
6.72/2.66	YES
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(47)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	The TRS R consists of the following rules:
6.72/2.66	
6.72/2.66	   active(f(0)) -> mark(cons(0, f(s(0))))
6.72/2.66	   active(f(s(0))) -> mark(f(p(s(0))))
6.72/2.66	   active(p(s(0))) -> mark(0)
6.72/2.66	   mark(f(X)) -> active(f(mark(X)))
6.72/2.66	   mark(0) -> active(0)
6.72/2.66	   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
6.72/2.66	   mark(s(X)) -> active(s(mark(X)))
6.72/2.66	   mark(p(X)) -> active(p(mark(X)))
6.72/2.66	   f(mark(X)) -> f(X)
6.72/2.66	   f(active(X)) -> f(X)
6.72/2.66	   cons(mark(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, mark(X2)) -> cons(X1, X2)
6.72/2.66	   cons(active(X1), X2) -> cons(X1, X2)
6.72/2.66	   cons(X1, active(X2)) -> cons(X1, X2)
6.72/2.66	   s(mark(X)) -> s(X)
6.72/2.66	   s(active(X)) -> s(X)
6.72/2.66	   p(mark(X)) -> p(X)
6.72/2.66	   p(active(X)) -> p(X)
6.72/2.66	
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(48) UsableRulesProof (EQUIVALENT)
6.72/2.66	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.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(49)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(50) QReductionProof (EQUIVALENT)
6.72/2.66	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.72/2.66	
6.72/2.66	   active(f(0))
6.72/2.66	   active(f(s(0)))
6.72/2.66	   active(p(s(0)))
6.72/2.66	   mark(f(x0))
6.72/2.66	   mark(0)
6.72/2.66	   mark(cons(x0, x1))
6.72/2.66	   mark(s(x0))
6.72/2.66	   mark(p(x0))
6.72/2.66	   f(mark(x0))
6.72/2.66	   f(active(x0))
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(51)
6.72/2.66	Obligation:
6.72/2.66	Q DP problem:
6.72/2.66	The TRS P consists of the following rules:
6.72/2.66	
6.72/2.66	   MARK(s(X)) -> MARK(X)
6.72/2.66	   MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	   MARK(p(X)) -> MARK(X)
6.72/2.66	
6.72/2.66	R is empty.
6.72/2.66	The set Q consists of the following terms:
6.72/2.66	
6.72/2.66	   cons(mark(x0), x1)
6.72/2.66	   cons(x0, mark(x1))
6.72/2.66	   cons(active(x0), x1)
6.72/2.66	   cons(x0, active(x1))
6.72/2.66	   s(mark(x0))
6.72/2.66	   s(active(x0))
6.72/2.66	   p(mark(x0))
6.72/2.66	   p(active(x0))
6.72/2.66	
6.72/2.66	We have to consider all minimal (P,Q,R)-chains.
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(52) QDPSizeChangeProof (EQUIVALENT)
6.72/2.66	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. 
6.72/2.66	
6.72/2.66	From the DPs we obtained the following set of size-change graphs:
6.72/2.66	*MARK(s(X)) -> MARK(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	*MARK(cons(X1, X2)) -> MARK(X1)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	*MARK(p(X)) -> MARK(X)
6.72/2.66	The graph contains the following edges 1 > 1
6.72/2.66	
6.72/2.66	
6.72/2.66	----------------------------------------
6.72/2.66	
6.72/2.66	(53)
6.72/2.66	YES
7.01/2.70	EOF