51.57/14.12	YES
51.86/14.15	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
51.86/14.15	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
51.86/14.15	
51.86/14.15	
51.86/14.15	Termination w.r.t. Q of the given QTRS could be proven:
51.86/14.15	
51.86/14.15	(0) QTRS
51.86/14.15	(1) QTRS Reverse [EQUIVALENT, 0 ms]
51.86/14.15	(2) QTRS
51.86/14.15	(3) DependencyPairsProof [EQUIVALENT, 213 ms]
51.86/14.15	(4) QDP
51.86/14.15	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
51.86/14.15	(6) AND
51.86/14.15	    (7) QDP
51.86/14.15	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
51.86/14.15	        (9) QDP
51.86/14.15	        (10) MRRProof [EQUIVALENT, 15 ms]
51.86/14.15	        (11) QDP
51.86/14.15	        (12) PisEmptyProof [EQUIVALENT, 0 ms]
51.86/14.15	        (13) YES
51.86/14.15	    (14) QDP
51.86/14.15	        (15) QDPOrderProof [EQUIVALENT, 859 ms]
51.86/14.15	        (16) QDP
51.86/14.15	        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
51.86/14.15	        (18) QDP
51.86/14.15	        (19) UsableRulesProof [EQUIVALENT, 0 ms]
51.86/14.15	        (20) QDP
51.86/14.15	        (21) QDPOrderProof [EQUIVALENT, 11 ms]
51.86/14.15	        (22) QDP
51.86/14.15	        (23) PisEmptyProof [EQUIVALENT, 0 ms]
51.86/14.15	        (24) YES
51.86/14.15	
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(0)
51.86/14.15	Obligation:
51.86/14.15	Q restricted rewrite system:
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   0(1(1(2(x1)))) -> 1(3(0(1(2(x1)))))
51.86/14.15	   0(1(1(2(x1)))) -> 1(3(1(0(2(x1)))))
51.86/14.15	   0(1(1(2(x1)))) -> 2(3(1(3(0(1(x1))))))
51.86/14.15	   0(1(2(0(x1)))) -> 1(0(0(2(2(x1)))))
51.86/14.15	   0(1(2(4(x1)))) -> 0(2(3(4(1(x1)))))
51.86/14.15	   0(1(2(4(x1)))) -> 1(0(4(2(3(x1)))))
51.86/14.15	   0(1(2(4(x1)))) -> 1(2(3(0(4(x1)))))
51.86/14.15	   0(1(2(4(x1)))) -> 2(1(4(0(3(x1)))))
51.86/14.15	   0(1(2(4(x1)))) -> 1(2(3(4(0(5(x1))))))
51.86/14.15	   0(1(2(4(x1)))) -> 2(1(1(0(3(4(x1))))))
51.86/14.15	   0(1(2(4(x1)))) -> 4(1(2(2(3(0(x1))))))
51.86/14.15	   0(1(2(4(x1)))) -> 4(5(0(2(3(1(x1))))))
51.86/14.15	   0(4(2(0(x1)))) -> 0(4(2(3(0(x1)))))
51.86/14.15	   0(4(2(0(x1)))) -> 2(3(0(4(0(0(x1))))))
51.86/14.15	   0(4(2(0(x1)))) -> 3(4(0(2(3(0(x1))))))
51.86/14.15	   0(4(2(0(x1)))) -> 4(2(3(0(4(0(x1))))))
51.86/14.15	   0(4(2(0(x1)))) -> 4(3(0(0(5(2(x1))))))
51.86/14.15	   0(4(2(4(x1)))) -> 4(0(2(3(1(4(x1))))))
51.86/14.15	   0(4(2(4(x1)))) -> 4(0(4(2(0(3(x1))))))
51.86/14.15	   5(1(2(0(x1)))) -> 2(1(3(5(0(x1)))))
51.86/14.15	   5(1(2(0(x1)))) -> 3(1(0(2(5(x1)))))
51.86/14.15	   5(1(2(0(x1)))) -> 2(3(0(0(1(5(x1))))))
51.86/14.15	   5(1(2(4(x1)))) -> 3(2(5(1(4(x1)))))
51.86/14.15	   5(1(2(4(x1)))) -> 5(2(1(3(4(x1)))))
51.86/14.15	   5(1(2(4(x1)))) -> 5(5(2(1(4(x1)))))
51.86/14.15	   5(1(2(4(x1)))) -> 0(3(4(5(2(1(x1))))))
51.86/14.15	   5(1(4(2(x1)))) -> 3(4(5(2(1(x1)))))
51.86/14.15	   5(1(4(2(x1)))) -> 2(1(3(4(5(4(x1))))))
51.86/14.15	   5(1(4(2(x1)))) -> 3(3(4(2(1(5(x1))))))
51.86/14.15	   5(4(1(4(x1)))) -> 3(4(4(1(5(4(x1))))))
51.86/14.15	   5(4(1(4(x1)))) -> 4(1(3(5(0(4(x1))))))
51.86/14.15	   5(4(1(4(x1)))) -> 5(1(3(4(5(4(x1))))))
51.86/14.15	   5(4(1(4(x1)))) -> 5(1(5(3(4(4(x1))))))
51.86/14.15	   5(4(2(0(x1)))) -> 5(4(2(3(0(x1)))))
51.86/14.15	   5(4(2(0(x1)))) -> 0(1(2(3(4(5(x1))))))
51.86/14.15	   5(4(2(0(x1)))) -> 4(5(3(3(0(2(x1))))))
51.86/14.15	   0(1(2(0(4(x1))))) -> 0(2(4(1(3(0(x1))))))
51.86/14.15	   0(1(2(0(4(x1))))) -> 2(0(3(4(0(1(x1))))))
51.86/14.15	   0(1(2(0(4(x1))))) -> 4(0(2(3(1(0(x1))))))
51.86/14.15	   0(1(4(2(2(x1))))) -> 1(2(3(4(0(2(x1))))))
51.86/14.15	   5(0(1(2(4(x1))))) -> 3(0(2(1(4(5(x1))))))
51.86/14.15	   5(1(2(2(4(x1))))) -> 5(2(2(1(4(5(x1))))))
51.86/14.15	   5(1(2(4(0(x1))))) -> 5(0(2(1(3(4(x1))))))
51.86/14.15	   5(1(2(4(0(x1))))) -> 5(0(3(2(1(4(x1))))))
51.86/14.15	   5(1(3(1(4(x1))))) -> 1(5(1(3(4(0(x1))))))
51.86/14.15	   5(1(4(1(2(x1))))) -> 2(1(5(3(4(1(x1))))))
51.86/14.15	   5(4(1(1(4(x1))))) -> 1(1(3(4(5(4(x1))))))
51.86/14.15	   5(4(1(4(0(x1))))) -> 3(4(0(4(5(1(x1))))))
51.86/14.15	   5(4(3(2(0(x1))))) -> 5(4(0(3(2(3(x1))))))
51.86/14.15	   5(4(5(2(0(x1))))) -> 5(0(5(4(4(2(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(1) QTRS Reverse (EQUIVALENT)
51.86/14.15	We applied the QTRS Reverse Processor [REVERSE].
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(2)
51.86/14.15	Obligation:
51.86/14.15	Q restricted rewrite system:
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(3) DependencyPairsProof (EQUIVALENT)
51.86/14.15	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(4)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(1(0(3(1(x1)))))
51.86/14.15	   2^1(1(1(0(x1)))) -> 0^1(3(1(x1)))
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(0(1(3(1(x1)))))
51.86/14.15	   2^1(1(1(0(x1)))) -> 0^1(1(3(1(x1))))
51.86/14.15	   2^1(1(1(0(x1)))) -> 0^1(3(1(3(2(x1)))))
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(x1)
51.86/14.15	   0^1(2(1(0(x1)))) -> 2^1(2(0(0(1(x1)))))
51.86/14.15	   0^1(2(1(0(x1)))) -> 2^1(0(0(1(x1))))
51.86/14.15	   0^1(2(1(0(x1)))) -> 0^1(0(1(x1)))
51.86/14.15	   0^1(2(1(0(x1)))) -> 0^1(1(x1))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(3(2(0(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(0(x1))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(4(0(1(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(0(1(x1)))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(1(x1))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(0(3(2(1(x1)))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(3(2(1(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(1(x1))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(4(1(2(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(x1)
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(4(3(2(1(x1)))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(3(2(1(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(3(0(1(1(2(x1))))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(1(1(2(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(3(2(2(1(4(x1))))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(2(1(4(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(1(4(x1)))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(1(0(x1)))) -> 2^1(0(5(4(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(5(4(x1)))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(3(2(4(0(x1)))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(0(4(0(3(2(x1))))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(4(0(3(2(x1)))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(0(3(2(x1))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(3(2(x1)))
51.86/14.15	   0^1(2(4(0(x1)))) -> 2^1(x1)
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(3(2(0(4(3(x1))))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 2^1(0(4(3(x1))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(4(3(x1)))
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(3(x1))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(4(0(3(2(4(x1))))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(0(3(2(4(x1)))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(3(2(4(x1))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 2^1(4(x1))
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   0^1(2(4(0(x1)))) -> 2^1(5(0(0(3(4(x1))))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(0(3(4(x1))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 0^1(3(4(x1)))
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(1(3(2(0(4(x1))))))
51.86/14.15	   4^1(2(4(0(x1)))) -> 2^1(0(4(x1)))
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(4(x1))
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(2(4(0(4(x1)))))
51.86/14.15	   4^1(2(4(0(x1)))) -> 2^1(4(0(4(x1))))
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(0(4(x1)))
51.86/14.15	   0^1(2(1(5(x1)))) -> 0^1(5(3(1(2(x1)))))
51.86/14.15	   0^1(2(1(5(x1)))) -> 2^1(x1)
51.86/14.15	   0^1(2(1(5(x1)))) -> 2^1(0(1(3(x1))))
51.86/14.15	   0^1(2(1(5(x1)))) -> 0^1(1(3(x1)))
51.86/14.15	   0^1(2(1(5(x1)))) -> 0^1(0(3(2(x1))))
51.86/14.15	   0^1(2(1(5(x1)))) -> 0^1(3(2(x1)))
51.86/14.15	   4^1(2(1(5(x1)))) -> 4^1(1(5(2(3(x1)))))
51.86/14.15	   4^1(2(1(5(x1)))) -> 2^1(3(x1))
51.86/14.15	   4^1(2(1(5(x1)))) -> 4^1(3(1(2(5(x1)))))
51.86/14.15	   4^1(2(1(5(x1)))) -> 2^1(5(x1))
51.86/14.15	   4^1(2(1(5(x1)))) -> 4^1(1(2(5(5(x1)))))
51.86/14.15	   4^1(2(1(5(x1)))) -> 2^1(5(5(x1)))
51.86/14.15	   4^1(2(1(5(x1)))) -> 2^1(5(4(3(0(x1)))))
51.86/14.15	   4^1(2(1(5(x1)))) -> 4^1(3(0(x1)))
51.86/14.15	   4^1(2(1(5(x1)))) -> 0^1(x1)
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(5(4(3(x1))))
51.86/14.15	   2^1(4(1(5(x1)))) -> 4^1(3(x1))
51.86/14.15	   2^1(4(1(5(x1)))) -> 4^1(5(4(3(1(2(x1))))))
51.86/14.15	   2^1(4(1(5(x1)))) -> 4^1(3(1(2(x1))))
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(4(3(3(x1))))
51.86/14.15	   2^1(4(1(5(x1)))) -> 4^1(3(3(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(5(1(4(4(3(x1))))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(4(3(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(3(x1))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(0(5(3(1(4(x1))))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 0^1(5(3(1(4(x1)))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(5(4(3(1(5(x1))))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(3(1(5(x1))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(4(3(5(1(5(x1))))))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(3(5(1(5(x1)))))
51.86/14.15	   0^1(2(4(5(x1)))) -> 0^1(3(2(4(5(x1)))))
51.86/14.15	   0^1(2(4(5(x1)))) -> 4^1(3(2(1(0(x1)))))
51.86/14.15	   0^1(2(4(5(x1)))) -> 2^1(1(0(x1)))
51.86/14.15	   0^1(2(4(5(x1)))) -> 0^1(x1)
51.86/14.15	   0^1(2(4(5(x1)))) -> 2^1(0(3(3(5(4(x1))))))
51.86/14.15	   0^1(2(4(5(x1)))) -> 0^1(3(3(5(4(x1)))))
51.86/14.15	   0^1(2(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(3(1(4(2(0(x1))))))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(2(0(x1)))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 2^1(0(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(4(3(0(2(x1)))))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(3(0(2(x1))))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(2(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 2^1(x1)
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(1(3(2(0(4(x1))))))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 2^1(0(4(x1)))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(4(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(x1)
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 2^1(0(4(3(2(1(x1))))))
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 0^1(4(3(2(1(x1)))))
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 4^1(3(2(1(x1))))
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 2^1(1(x1))
51.86/14.15	   4^1(2(1(0(5(x1))))) -> 4^1(1(2(0(3(x1)))))
51.86/14.15	   4^1(2(1(0(5(x1))))) -> 2^1(0(3(x1)))
51.86/14.15	   4^1(2(1(0(5(x1))))) -> 0^1(3(x1))
51.86/14.15	   4^1(2(2(1(5(x1))))) -> 4^1(1(2(2(5(x1)))))
51.86/14.15	   4^1(2(2(1(5(x1))))) -> 2^1(2(5(x1)))
51.86/14.15	   4^1(2(2(1(5(x1))))) -> 2^1(5(x1))
51.86/14.15	   0^1(4(2(1(5(x1))))) -> 4^1(3(1(2(0(5(x1))))))
51.86/14.15	   0^1(4(2(1(5(x1))))) -> 2^1(0(5(x1)))
51.86/14.15	   0^1(4(2(1(5(x1))))) -> 0^1(5(x1))
51.86/14.15	   0^1(4(2(1(5(x1))))) -> 4^1(1(2(3(0(5(x1))))))
51.86/14.15	   0^1(4(2(1(5(x1))))) -> 2^1(3(0(5(x1))))
51.86/14.15	   4^1(1(3(1(5(x1))))) -> 0^1(4(3(1(5(1(x1))))))
51.86/14.15	   4^1(1(3(1(5(x1))))) -> 4^1(3(1(5(1(x1)))))
51.86/14.15	   2^1(1(4(1(5(x1))))) -> 4^1(3(5(1(2(x1)))))
51.86/14.15	   2^1(1(4(1(5(x1))))) -> 2^1(x1)
51.86/14.15	   4^1(1(1(4(5(x1))))) -> 4^1(5(4(3(1(1(x1))))))
51.86/14.15	   4^1(1(1(4(5(x1))))) -> 4^1(3(1(1(x1))))
51.86/14.15	   0^1(4(1(4(5(x1))))) -> 4^1(0(4(3(x1))))
51.86/14.15	   0^1(4(1(4(5(x1))))) -> 0^1(4(3(x1)))
51.86/14.15	   0^1(4(1(4(5(x1))))) -> 4^1(3(x1))
51.86/14.15	   0^1(2(3(4(5(x1))))) -> 2^1(3(0(4(5(x1)))))
51.86/14.15	   0^1(2(3(4(5(x1))))) -> 0^1(4(5(x1)))
51.86/14.15	   0^1(2(5(4(5(x1))))) -> 2^1(4(4(5(0(5(x1))))))
51.86/14.15	   0^1(2(5(4(5(x1))))) -> 4^1(4(5(0(5(x1)))))
51.86/14.15	   0^1(2(5(4(5(x1))))) -> 4^1(5(0(5(x1))))
51.86/14.15	   0^1(2(5(4(5(x1))))) -> 0^1(5(x1))
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(5) DependencyGraphProof (EQUIVALENT)
51.86/14.15	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 116 less nodes.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(6)
51.86/14.15	Complex Obligation (AND)
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(7)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 2^1(1(x1))
51.86/14.15	   2^1(1(4(1(5(x1))))) -> 2^1(x1)
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(8) UsableRulesProof (EQUIVALENT)
51.86/14.15	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.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(9)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 2^1(1(x1))
51.86/14.15	   2^1(1(4(1(5(x1))))) -> 2^1(x1)
51.86/14.15	
51.86/14.15	R is empty.
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(10) MRRProof (EQUIVALENT)
51.86/14.15	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.
51.86/14.15	
51.86/14.15	Strictly oriented dependency pairs:
51.86/14.15	
51.86/14.15	   2^1(4(1(5(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(1(1(0(x1)))) -> 2^1(x1)
51.86/14.15	   2^1(2(4(1(0(x1))))) -> 2^1(1(x1))
51.86/14.15	   2^1(1(4(1(5(x1))))) -> 2^1(x1)
51.86/14.15	
51.86/14.15	
51.86/14.15	Used ordering: Polynomial interpretation [POLO]:
51.86/14.15	
51.86/14.15	   POL(0(x_1)) = x_1
51.86/14.15	   POL(1(x_1)) = 1 + x_1
51.86/14.15	   POL(2(x_1)) = 2 + 2*x_1
51.86/14.15	   POL(2^1(x_1)) = x_1
51.86/14.15	   POL(4(x_1)) = x_1
51.86/14.15	   POL(5(x_1)) = 3 + 2*x_1
51.86/14.15	
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(11)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	P is empty.
51.86/14.15	R is empty.
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(12) PisEmptyProof (EQUIVALENT)
51.86/14.15	The TRS P is empty. Hence, there is no (P,Q,R) chain.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(13)
51.86/14.15	YES
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(14)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(4(1(2(x1))))
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(4(x1))
51.86/14.15	   0^1(2(4(5(x1)))) -> 0^1(x1)
51.86/14.15	   0^1(2(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(2(4(0(4(x1)))))
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(0(4(x1)))
51.86/14.15	   4^1(2(1(5(x1)))) -> 0^1(x1)
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(2(0(x1)))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(2(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(4(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(x1)
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(15) QDPOrderProof (EQUIVALENT)
51.86/14.15	We use the reduction pair processor [LPAR04,JAR06].
51.86/14.15	
51.86/14.15	
51.86/14.15	The following pairs can be oriented strictly and are deleted.
51.86/14.15	
51.86/14.15	   0^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(4(x1))
51.86/14.15	   0^1(2(4(5(x1)))) -> 0^1(x1)
51.86/14.15	   0^1(2(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 4^1(0(4(x1)))
51.86/14.15	   4^1(2(1(5(x1)))) -> 0^1(x1)
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(2(0(x1)))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(2(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 0^1(4(x1))
51.86/14.15	   4^1(0(2(1(0(x1))))) -> 4^1(x1)
51.86/14.15	The remaining pairs can at least be oriented weakly.
51.86/14.15	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
51.86/14.15	
51.86/14.15	POL( 0^1_1(x_1) ) = 2x_1 + 2
51.86/14.15	POL( 4^1_1(x_1) ) = 2x_1 + 2
51.86/14.15	POL( 2_1(x_1) ) = x_1 + 2
51.86/14.15	POL( 1_1(x_1) ) = x_1
51.86/14.15	POL( 0_1(x_1) ) = 2x_1
51.86/14.15	POL( 3_1(x_1) ) = max{0, -2}
51.86/14.15	POL( 4_1(x_1) ) = x_1
51.86/14.15	POL( 5_1(x_1) ) = x_1
51.86/14.15	
51.86/14.15	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(16)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   4^1(2(1(0(x1)))) -> 0^1(4(1(2(x1))))
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	   4^1(2(4(0(x1)))) -> 0^1(2(4(0(4(x1)))))
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(17) DependencyGraphProof (EQUIVALENT)
51.86/14.15	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(18)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   0(2(1(0(x1)))) -> 2(2(0(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(4(3(2(0(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(2(4(0(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(0(3(2(1(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 3(0(4(1(2(x1)))))
51.86/14.15	   4(2(1(0(x1)))) -> 5(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 4(3(0(1(1(2(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 0(3(2(2(1(4(x1))))))
51.86/14.15	   4(2(1(0(x1)))) -> 1(3(2(0(5(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(4(0(x1)))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(0(4(0(3(2(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(3(2(0(4(3(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 0(4(0(3(2(4(x1))))))
51.86/14.15	   0(2(4(0(x1)))) -> 2(5(0(0(3(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 4(1(3(2(0(4(x1))))))
51.86/14.15	   4(2(4(0(x1)))) -> 3(0(2(4(0(4(x1))))))
51.86/14.15	   0(2(1(5(x1)))) -> 0(5(3(1(2(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(2(0(1(3(x1)))))
51.86/14.15	   0(2(1(5(x1)))) -> 5(1(0(0(3(2(x1))))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(5(2(3(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(3(1(2(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 4(1(2(5(5(x1)))))
51.86/14.15	   4(2(1(5(x1)))) -> 1(2(5(4(3(0(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(1(4(4(3(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(0(5(3(1(4(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(5(4(3(1(5(x1))))))
51.86/14.15	   4(1(4(5(x1)))) -> 4(4(3(5(1(5(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 0(3(2(4(5(x1)))))
51.86/14.15	   0(2(4(5(x1)))) -> 5(4(3(2(1(0(x1))))))
51.86/14.15	   0(2(4(5(x1)))) -> 2(0(3(3(5(4(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(3(1(4(2(0(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 1(0(4(3(0(2(x1))))))
51.86/14.15	   4(0(2(1(0(x1))))) -> 0(1(3(2(0(4(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   4(2(1(0(5(x1))))) -> 5(4(1(2(0(3(x1))))))
51.86/14.15	   4(2(2(1(5(x1))))) -> 5(4(1(2(2(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(3(1(2(0(5(x1))))))
51.86/14.15	   0(4(2(1(5(x1))))) -> 4(1(2(3(0(5(x1))))))
51.86/14.15	   4(1(3(1(5(x1))))) -> 0(4(3(1(5(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	   4(1(1(4(5(x1))))) -> 4(5(4(3(1(1(x1))))))
51.86/14.15	   0(4(1(4(5(x1))))) -> 1(5(4(0(4(3(x1))))))
51.86/14.15	   0(2(3(4(5(x1))))) -> 3(2(3(0(4(5(x1))))))
51.86/14.15	   0(2(5(4(5(x1))))) -> 2(4(4(5(0(5(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(19) UsableRulesProof (EQUIVALENT)
51.86/14.15	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.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(20)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	The TRS P consists of the following rules:
51.86/14.15	
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(21) QDPOrderProof (EQUIVALENT)
51.86/14.15	We use the reduction pair processor [LPAR04,JAR06].
51.86/14.15	
51.86/14.15	
51.86/14.15	The following pairs can be oriented strictly and are deleted.
51.86/14.15	
51.86/14.15	   4^1(2(1(0(x1)))) -> 4^1(1(2(x1)))
51.86/14.15	   4^1(1(4(5(x1)))) -> 4^1(x1)
51.86/14.15	The remaining pairs can at least be oriented weakly.
51.86/14.15	Used ordering:  Polynomial interpretation [POLO]:
51.86/14.15	
51.86/14.15	   POL(0(x_1)) = 1 + x_1
51.86/14.15	   POL(1(x_1)) = x_1
51.86/14.15	   POL(2(x_1)) = 1 + x_1
51.86/14.15	   POL(3(x_1)) = 0
51.86/14.15	   POL(4(x_1)) = x_1
51.86/14.15	   POL(4^1(x_1)) = x_1
51.86/14.15	   POL(5(x_1)) = 1 + x_1
51.86/14.15	
51.86/14.15	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	
51.86/14.15	
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(22)
51.86/14.15	Obligation:
51.86/14.15	Q DP problem:
51.86/14.15	P is empty.
51.86/14.15	The TRS R consists of the following rules:
51.86/14.15	
51.86/14.15	   2(1(1(0(x1)))) -> 2(1(0(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 2(0(1(3(1(x1)))))
51.86/14.15	   2(1(1(0(x1)))) -> 1(0(3(1(3(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 1(2(5(4(3(x1)))))
51.86/14.15	   2(4(1(5(x1)))) -> 4(5(4(3(1(2(x1))))))
51.86/14.15	   2(4(1(5(x1)))) -> 5(1(2(4(3(3(x1))))))
51.86/14.15	   2(2(4(1(0(x1))))) -> 2(0(4(3(2(1(x1))))))
51.86/14.15	   2(1(4(1(5(x1))))) -> 1(4(3(5(1(2(x1))))))
51.86/14.15	
51.86/14.15	Q is empty.
51.86/14.15	We have to consider all minimal (P,Q,R)-chains.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(23) PisEmptyProof (EQUIVALENT)
51.86/14.15	The TRS P is empty. Hence, there is no (P,Q,R) chain.
51.86/14.15	----------------------------------------
51.86/14.15	
51.86/14.15	(24)
51.86/14.15	YES
52.03/14.26	EOF