20.18/5.97	YES
20.18/6.03	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
20.18/6.03	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
20.18/6.03	
20.18/6.03	
20.18/6.03	Termination w.r.t. Q of the given QTRS could be proven:
20.18/6.03	
20.18/6.03	(0) QTRS
20.18/6.03	(1) DependencyPairsProof [EQUIVALENT, 197 ms]
20.18/6.03	(2) QDP
20.18/6.03	(3) DependencyGraphProof [EQUIVALENT, 3 ms]
20.18/6.03	(4) AND
20.18/6.03	    (5) QDP
20.18/6.03	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
20.18/6.03	        (7) QDP
20.18/6.03	        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
20.18/6.03	        (9) YES
20.18/6.03	    (10) QDP
20.18/6.03	        (11) UsableRulesProof [EQUIVALENT, 0 ms]
20.18/6.03	        (12) QDP
20.18/6.03	        (13) QDPOrderProof [EQUIVALENT, 15 ms]
20.18/6.03	        (14) QDP
20.18/6.03	        (15) PisEmptyProof [EQUIVALENT, 0 ms]
20.18/6.03	        (16) YES
20.18/6.03	
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(0)
20.18/6.03	Obligation:
20.18/6.03	Q restricted rewrite system:
20.18/6.03	The TRS R consists of the following rules:
20.18/6.03	
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
20.18/6.03	   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
20.18/6.03	   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
20.18/6.03	   0(4(1(x1))) -> 0(1(4(4(x1))))
20.18/6.03	   0(4(1(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(4(3(x1))))
20.18/6.03	   2(0(1(x1))) -> 5(0(2(1(x1))))
20.18/6.03	   2(3(1(x1))) -> 1(3(5(2(x1))))
20.18/6.03	   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
20.18/6.03	   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
20.18/6.03	   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
20.18/6.03	   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
20.18/6.03	   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
20.18/6.03	   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
20.18/6.03	   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
20.18/6.03	   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
20.18/6.03	   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
20.18/6.03	   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
20.18/6.03	   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
20.18/6.03	   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
20.18/6.03	   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
20.18/6.03	   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
20.18/6.03	   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
20.18/6.03	   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
20.18/6.03	   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
20.18/6.03	   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
20.18/6.03	   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
20.18/6.03	   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
20.18/6.03	   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
20.18/6.03	   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))
20.18/6.03	
20.18/6.03	Q is empty.
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(1) DependencyPairsProof (EQUIVALENT)
20.18/6.03	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(2)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	The TRS P consists of the following rules:
20.18/6.03	
20.18/6.03	   0^1(1(2(x1))) -> 0^1(0(2(1(x1))))
20.18/6.03	   0^1(1(2(x1))) -> 0^1(2(1(x1)))
20.18/6.03	   0^1(1(2(x1))) -> 2^1(1(x1))
20.18/6.03	   0^1(1(2(x1))) -> 0^1(2(1(3(x1))))
20.18/6.03	   0^1(1(2(x1))) -> 2^1(1(3(x1)))
20.18/6.03	   0^1(1(2(x1))) -> 0^1(0(2(1(4(4(x1))))))
20.18/6.03	   0^1(1(2(x1))) -> 0^1(2(1(4(4(x1)))))
20.18/6.03	   0^1(1(2(x1))) -> 2^1(1(4(4(x1))))
20.18/6.03	   0^1(3(1(x1))) -> 0^1(1(3(4(0(x1)))))
20.18/6.03	   0^1(3(1(x1))) -> 0^1(x1)
20.18/6.03	   0^1(3(1(x1))) -> 0^1(1(3(4(4(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(1(3(x1))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(1(3(x1)))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(3(4(x1))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(3(4(x1)))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(0(2(4(3(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(4(3(x1))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(4(3(x1)))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(1(4(3(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(1(4(3(x1))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(4(3(3(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(4(3(3(x1))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(1(3(3(4(x1))))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(1(3(3(4(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(2(3(4(5(5(x1))))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(3(4(5(5(x1)))))
20.18/6.03	   0^1(3(2(x1))) -> 2^1(4(4(3(4(0(x1))))))
20.18/6.03	   0^1(3(2(x1))) -> 0^1(x1)
20.18/6.03	   0^1(4(1(x1))) -> 0^1(1(4(4(x1))))
20.18/6.03	   0^1(4(1(x1))) -> 0^1(2(1(4(x1))))
20.18/6.03	   0^1(4(1(x1))) -> 2^1(1(4(x1)))
20.18/6.03	   0^1(4(2(x1))) -> 0^1(2(1(4(x1))))
20.18/6.03	   0^1(4(2(x1))) -> 2^1(1(4(x1)))
20.18/6.03	   0^1(4(2(x1))) -> 0^1(2(3(4(x1))))
20.18/6.03	   0^1(4(2(x1))) -> 2^1(3(4(x1)))
20.18/6.03	   0^1(4(2(x1))) -> 0^1(2(4(3(x1))))
20.18/6.03	   0^1(4(2(x1))) -> 2^1(4(3(x1)))
20.18/6.03	   2^1(0(1(x1))) -> 0^1(2(1(x1)))
20.18/6.03	   2^1(0(1(x1))) -> 2^1(1(x1))
20.18/6.03	   2^1(3(1(x1))) -> 2^1(x1)
20.18/6.03	   2^1(3(1(x1))) -> 0^1(2(1(3(5(x1)))))
20.18/6.03	   2^1(3(1(x1))) -> 2^1(1(3(5(x1))))
20.18/6.03	   0^1(2(0(1(x1)))) -> 0^1(0(2(1(x1))))
20.18/6.03	   0^1(2(0(1(x1)))) -> 0^1(2(1(x1)))
20.18/6.03	   0^1(2(0(1(x1)))) -> 2^1(1(x1))
20.18/6.03	   0^1(3(1(1(x1)))) -> 0^1(1(4(1(3(4(x1))))))
20.18/6.03	   0^1(3(2(1(x1)))) -> 0^1(0(3(4(2(1(x1))))))
20.18/6.03	   0^1(3(2(1(x1)))) -> 0^1(3(4(2(1(x1)))))
20.18/6.03	   0^1(3(2(2(x1)))) -> 0^1(2(2(x1)))
20.18/6.03	   0^1(4(1(2(x1)))) -> 0^1(2(5(x1)))
20.18/6.03	   0^1(4(1(2(x1)))) -> 2^1(5(x1))
20.18/6.03	   0^1(4(3(2(x1)))) -> 2^1(3(4(4(0(0(x1))))))
20.18/6.03	   0^1(4(3(2(x1)))) -> 0^1(0(x1))
20.18/6.03	   0^1(4(3(2(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(1(4(3(5(4(x1))))))
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(1(5(3(4(0(x1))))))
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(5(3(2(x1)))) -> 0^1(2(4(5(3(x1)))))
20.18/6.03	   0^1(5(3(2(x1)))) -> 2^1(4(5(3(x1))))
20.18/6.03	   0^1(5(3(2(x1)))) -> 0^1(2(5(3(3(x1)))))
20.18/6.03	   0^1(5(3(2(x1)))) -> 2^1(5(3(3(x1))))
20.18/6.03	   2^1(0(3(1(x1)))) -> 2^1(0(1(3(5(2(x1))))))
20.18/6.03	   2^1(0(3(1(x1)))) -> 0^1(1(3(5(2(x1)))))
20.18/6.03	   2^1(0(3(1(x1)))) -> 2^1(x1)
20.18/6.03	   2^1(0(4(1(x1)))) -> 2^1(0(1(4(5(x1)))))
20.18/6.03	   2^1(0(4(1(x1)))) -> 0^1(1(4(5(x1))))
20.18/6.03	   2^1(5(3(2(x1)))) -> 2^1(5(2(3(3(x1)))))
20.18/6.03	   2^1(5(3(2(x1)))) -> 2^1(3(3(x1)))
20.18/6.03	   2^1(5(4(2(x1)))) -> 0^1(2(5(2(4(x1)))))
20.18/6.03	   2^1(5(4(2(x1)))) -> 2^1(5(2(4(x1))))
20.18/6.03	   2^1(5(4(2(x1)))) -> 2^1(4(x1))
20.18/6.03	   0^1(0(3(2(1(x1))))) -> 0^1(0(1(3(5(2(x1))))))
20.18/6.03	   0^1(0(3(2(1(x1))))) -> 0^1(1(3(5(2(x1)))))
20.18/6.03	   0^1(0(3(2(1(x1))))) -> 2^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(1(4(3(2(0(x1))))))
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 2^1(0(x1))
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 2^1(3(1(0(0(5(x1))))))
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(0(5(x1)))
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
20.18/6.03	   0^1(3(2(5(1(x1))))) -> 0^1(2(5(1(3(3(x1))))))
20.18/6.03	   0^1(3(2(5(1(x1))))) -> 2^1(5(1(3(3(x1)))))
20.18/6.03	   0^1(5(1(1(2(x1))))) -> 0^1(2(4(1(1(5(x1))))))
20.18/6.03	   0^1(5(1(1(2(x1))))) -> 2^1(4(1(1(5(x1)))))
20.18/6.03	   0^1(5(1(2(2(x1))))) -> 0^1(2(5(2(1(2(x1))))))
20.18/6.03	   0^1(5(1(2(2(x1))))) -> 2^1(5(2(1(2(x1)))))
20.18/6.03	   0^1(5(1(2(2(x1))))) -> 2^1(1(2(x1)))
20.18/6.03	   0^1(5(3(2(1(x1))))) -> 0^1(1(3(4(2(5(x1))))))
20.18/6.03	   0^1(5(3(2(1(x1))))) -> 2^1(5(x1))
20.18/6.03	   0^1(5(5(3(2(x1))))) -> 0^1(2(5(1(3(5(x1))))))
20.18/6.03	   0^1(5(5(3(2(x1))))) -> 2^1(5(1(3(5(x1)))))
20.18/6.03	   2^1(0(3(1(1(x1))))) -> 2^1(1(0(1(3(4(x1))))))
20.18/6.03	   2^1(0(3(1(1(x1))))) -> 0^1(1(3(4(x1))))
20.18/6.03	   2^1(2(0(3(1(x1))))) -> 0^1(2(5(2(x1))))
20.18/6.03	   2^1(2(0(3(1(x1))))) -> 2^1(5(2(x1)))
20.18/6.03	   2^1(2(0(3(1(x1))))) -> 2^1(x1)
20.18/6.03	   2^1(2(0(5(1(x1))))) -> 2^1(0(2(1(5(1(x1))))))
20.18/6.03	   2^1(2(0(5(1(x1))))) -> 0^1(2(1(5(1(x1)))))
20.18/6.03	   2^1(2(0(5(1(x1))))) -> 2^1(1(5(1(x1))))
20.18/6.03	   2^1(5(5(4(1(x1))))) -> 2^1(1(3(4(x1))))
20.18/6.03	
20.18/6.03	The TRS R consists of the following rules:
20.18/6.03	
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
20.18/6.03	   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
20.18/6.03	   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
20.18/6.03	   0(4(1(x1))) -> 0(1(4(4(x1))))
20.18/6.03	   0(4(1(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(4(3(x1))))
20.18/6.03	   2(0(1(x1))) -> 5(0(2(1(x1))))
20.18/6.03	   2(3(1(x1))) -> 1(3(5(2(x1))))
20.18/6.03	   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
20.18/6.03	   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
20.18/6.03	   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
20.18/6.03	   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
20.18/6.03	   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
20.18/6.03	   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
20.18/6.03	   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
20.18/6.03	   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
20.18/6.03	   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
20.18/6.03	   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
20.18/6.03	   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
20.18/6.03	   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
20.18/6.03	   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
20.18/6.03	   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
20.18/6.03	   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
20.18/6.03	   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
20.18/6.03	   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
20.18/6.03	   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
20.18/6.03	   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
20.18/6.03	   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
20.18/6.03	   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
20.18/6.03	   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))
20.18/6.03	
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(3) DependencyGraphProof (EQUIVALENT)
20.18/6.03	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 91 less nodes.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(4)
20.18/6.03	Complex Obligation (AND)
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(5)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	The TRS P consists of the following rules:
20.18/6.03	
20.18/6.03	   2^1(0(3(1(x1)))) -> 2^1(x1)
20.18/6.03	   2^1(3(1(x1))) -> 2^1(x1)
20.18/6.03	   2^1(2(0(3(1(x1))))) -> 2^1(x1)
20.18/6.03	
20.18/6.03	The TRS R consists of the following rules:
20.18/6.03	
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
20.18/6.03	   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
20.18/6.03	   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
20.18/6.03	   0(4(1(x1))) -> 0(1(4(4(x1))))
20.18/6.03	   0(4(1(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(4(3(x1))))
20.18/6.03	   2(0(1(x1))) -> 5(0(2(1(x1))))
20.18/6.03	   2(3(1(x1))) -> 1(3(5(2(x1))))
20.18/6.03	   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
20.18/6.03	   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
20.18/6.03	   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
20.18/6.03	   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
20.18/6.03	   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
20.18/6.03	   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
20.18/6.03	   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
20.18/6.03	   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
20.18/6.03	   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
20.18/6.03	   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
20.18/6.03	   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
20.18/6.03	   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
20.18/6.03	   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
20.18/6.03	   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
20.18/6.03	   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
20.18/6.03	   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
20.18/6.03	   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
20.18/6.03	   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
20.18/6.03	   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
20.18/6.03	   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
20.18/6.03	   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
20.18/6.03	   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))
20.18/6.03	
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(6) UsableRulesProof (EQUIVALENT)
20.18/6.03	We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(7)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	The TRS P consists of the following rules:
20.18/6.03	
20.18/6.03	   2^1(0(3(1(x1)))) -> 2^1(x1)
20.18/6.03	   2^1(3(1(x1))) -> 2^1(x1)
20.18/6.03	   2^1(2(0(3(1(x1))))) -> 2^1(x1)
20.18/6.03	
20.18/6.03	R is empty.
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(8) QDPSizeChangeProof (EQUIVALENT)
20.18/6.03	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. 
20.18/6.03	
20.18/6.03	From the DPs we obtained the following set of size-change graphs:
20.18/6.03	*2^1(0(3(1(x1)))) -> 2^1(x1)
20.18/6.03	The graph contains the following edges 1 > 1
20.18/6.03	
20.18/6.03	
20.18/6.03	*2^1(3(1(x1))) -> 2^1(x1)
20.18/6.03	The graph contains the following edges 1 > 1
20.18/6.03	
20.18/6.03	
20.18/6.03	*2^1(2(0(3(1(x1))))) -> 2^1(x1)
20.18/6.03	The graph contains the following edges 1 > 1
20.18/6.03	
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(9)
20.18/6.03	YES
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(10)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	The TRS P consists of the following rules:
20.18/6.03	
20.18/6.03	   0^1(3(2(x1))) -> 0^1(x1)
20.18/6.03	   0^1(3(1(x1))) -> 0^1(x1)
20.18/6.03	   0^1(4(3(2(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
20.18/6.03	
20.18/6.03	The TRS R consists of the following rules:
20.18/6.03	
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
20.18/6.03	   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
20.18/6.03	   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
20.18/6.03	   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
20.18/6.03	   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
20.18/6.03	   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
20.18/6.03	   0(4(1(x1))) -> 0(1(4(4(x1))))
20.18/6.03	   0(4(1(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(1(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(3(4(x1))))
20.18/6.03	   0(4(2(x1))) -> 0(2(4(3(x1))))
20.18/6.03	   2(0(1(x1))) -> 5(0(2(1(x1))))
20.18/6.03	   2(3(1(x1))) -> 1(3(5(2(x1))))
20.18/6.03	   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
20.18/6.03	   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
20.18/6.03	   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
20.18/6.03	   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
20.18/6.03	   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
20.18/6.03	   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
20.18/6.03	   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
20.18/6.03	   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
20.18/6.03	   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
20.18/6.03	   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
20.18/6.03	   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
20.18/6.03	   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
20.18/6.03	   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
20.18/6.03	   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
20.18/6.03	   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
20.18/6.03	   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
20.18/6.03	   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
20.18/6.03	   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
20.18/6.03	   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
20.18/6.03	   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
20.18/6.03	   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
20.18/6.03	   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
20.18/6.03	   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
20.18/6.03	   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
20.18/6.03	   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))
20.18/6.03	
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(11) UsableRulesProof (EQUIVALENT)
20.18/6.03	We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(12)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	The TRS P consists of the following rules:
20.18/6.03	
20.18/6.03	   0^1(3(2(x1))) -> 0^1(x1)
20.18/6.03	   0^1(3(1(x1))) -> 0^1(x1)
20.18/6.03	   0^1(4(3(2(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
20.18/6.03	
20.18/6.03	R is empty.
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(13) QDPOrderProof (EQUIVALENT)
20.18/6.03	We use the reduction pair processor [LPAR04,JAR06].
20.18/6.03	
20.18/6.03	
20.18/6.03	The following pairs can be oriented strictly and are deleted.
20.18/6.03	
20.18/6.03	   0^1(3(2(x1))) -> 0^1(x1)
20.18/6.03	   0^1(3(1(x1))) -> 0^1(x1)
20.18/6.03	   0^1(4(3(2(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(5(3(1(x1)))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(x1)
20.18/6.03	   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
20.18/6.03	The remaining pairs can at least be oriented weakly.
20.18/6.03	Used ordering:  Polynomial interpretation [POLO]:
20.18/6.03	
20.18/6.03	   POL(0(x_1)) = 1 + x_1
20.18/6.03	   POL(0^1(x_1)) = x_1
20.18/6.03	   POL(1(x_1)) = 1 + x_1
20.18/6.03	   POL(2(x_1)) = 1 + x_1
20.18/6.03	   POL(3(x_1)) = 1 + x_1
20.18/6.03	   POL(4(x_1)) = 1 + x_1
20.18/6.03	   POL(5(x_1)) = x_1
20.18/6.03	
20.18/6.03	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
20.18/6.03	none
20.18/6.03	
20.18/6.03	
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(14)
20.18/6.03	Obligation:
20.18/6.03	Q DP problem:
20.18/6.03	P is empty.
20.18/6.03	R is empty.
20.18/6.03	Q is empty.
20.18/6.03	We have to consider all minimal (P,Q,R)-chains.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(15) PisEmptyProof (EQUIVALENT)
20.18/6.03	The TRS P is empty. Hence, there is no (P,Q,R) chain.
20.18/6.03	----------------------------------------
20.18/6.03	
20.18/6.03	(16)
20.18/6.03	YES
20.50/10.16	EOF