6.31/2.54	NO
6.64/2.65	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
6.64/2.65	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
6.64/2.65	
6.64/2.65	
6.64/2.65	Termination w.r.t. Q of the given QTRS could be disproven:
6.64/2.65	
6.64/2.65	(0) QTRS
6.64/2.65	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
6.64/2.65	(2) QDP
6.64/2.65	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
6.64/2.65	(4) QDP
6.64/2.65	(5) UsableRulesProof [EQUIVALENT, 1 ms]
6.64/2.65	(6) QDP
6.64/2.65	(7) QReductionProof [EQUIVALENT, 0 ms]
6.64/2.65	(8) QDP
6.64/2.65	(9) NonTerminationLoopProof [COMPLETE, 0 ms]
6.64/2.65	(10) NO
6.64/2.65	
6.64/2.65	
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(0)
6.64/2.65	Obligation:
6.64/2.65	Q restricted rewrite system:
6.64/2.65	The TRS R consists of the following rules:
6.64/2.65	
6.64/2.65	   h(f(f(x))) -> h(f(g(f(x))))
6.64/2.65	   f(g(f(x))) -> f(f(x))
6.64/2.65	
6.64/2.65	The set Q consists of the following terms:
6.64/2.65	
6.64/2.65	   h(f(f(x0)))
6.64/2.65	   f(g(f(x0)))
6.64/2.65	
6.64/2.65	
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(1) DependencyPairsProof (EQUIVALENT)
6.64/2.65	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(2)
6.64/2.65	Obligation:
6.64/2.65	Q DP problem:
6.64/2.65	The TRS P consists of the following rules:
6.64/2.65	
6.64/2.65	   H(f(f(x))) -> H(f(g(f(x))))
6.64/2.65	   H(f(f(x))) -> F(g(f(x)))
6.64/2.65	   F(g(f(x))) -> F(f(x))
6.64/2.65	
6.64/2.65	The TRS R consists of the following rules:
6.64/2.65	
6.64/2.65	   h(f(f(x))) -> h(f(g(f(x))))
6.64/2.65	   f(g(f(x))) -> f(f(x))
6.64/2.65	
6.64/2.65	The set Q consists of the following terms:
6.64/2.65	
6.64/2.65	   h(f(f(x0)))
6.64/2.65	   f(g(f(x0)))
6.64/2.65	
6.64/2.65	We have to consider all minimal (P,Q,R)-chains.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(3) DependencyGraphProof (EQUIVALENT)
6.64/2.65	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(4)
6.64/2.65	Obligation:
6.64/2.65	Q DP problem:
6.64/2.65	The TRS P consists of the following rules:
6.64/2.65	
6.64/2.65	   H(f(f(x))) -> H(f(g(f(x))))
6.64/2.65	
6.64/2.65	The TRS R consists of the following rules:
6.64/2.65	
6.64/2.65	   h(f(f(x))) -> h(f(g(f(x))))
6.64/2.65	   f(g(f(x))) -> f(f(x))
6.64/2.65	
6.64/2.65	The set Q consists of the following terms:
6.64/2.65	
6.64/2.65	   h(f(f(x0)))
6.64/2.65	   f(g(f(x0)))
6.64/2.65	
6.64/2.65	We have to consider all minimal (P,Q,R)-chains.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(5) UsableRulesProof (EQUIVALENT)
6.64/2.65	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.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(6)
6.64/2.65	Obligation:
6.64/2.65	Q DP problem:
6.64/2.65	The TRS P consists of the following rules:
6.64/2.65	
6.64/2.65	   H(f(f(x))) -> H(f(g(f(x))))
6.64/2.65	
6.64/2.65	The TRS R consists of the following rules:
6.64/2.65	
6.64/2.65	   f(g(f(x))) -> f(f(x))
6.64/2.65	
6.64/2.65	The set Q consists of the following terms:
6.64/2.65	
6.64/2.65	   h(f(f(x0)))
6.64/2.65	   f(g(f(x0)))
6.64/2.65	
6.64/2.65	We have to consider all minimal (P,Q,R)-chains.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(7) QReductionProof (EQUIVALENT)
6.64/2.65	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
6.64/2.65	
6.64/2.65	   h(f(f(x0)))
6.64/2.65	
6.64/2.65	
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(8)
6.64/2.65	Obligation:
6.64/2.65	Q DP problem:
6.64/2.65	The TRS P consists of the following rules:
6.64/2.65	
6.64/2.65	   H(f(f(x))) -> H(f(g(f(x))))
6.64/2.65	
6.64/2.65	The TRS R consists of the following rules:
6.64/2.65	
6.64/2.65	   f(g(f(x))) -> f(f(x))
6.64/2.65	
6.64/2.65	The set Q consists of the following terms:
6.64/2.65	
6.64/2.65	   f(g(f(x0)))
6.64/2.65	
6.64/2.65	We have to consider all minimal (P,Q,R)-chains.
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(9) NonTerminationLoopProof (COMPLETE)
6.64/2.65	We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
6.64/2.65	Found a loop by narrowing to the left:
6.64/2.65	
6.64/2.65	s = H(f(g(f(x')))) evaluates to  t =H(f(g(f(x'))))
6.64/2.65	
6.64/2.65	Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
6.64/2.65	* Matcher: [ ]
6.64/2.65	* Semiunifier: [ ]
6.64/2.65	
6.64/2.65	--------------------------------------------------------------------------------
6.64/2.65	Rewriting sequence
6.64/2.65	
6.64/2.65	H(f(g(f(x')))) -> H(f(f(x')))
6.64/2.65	with rule f(g(f(x''))) -> f(f(x'')) at position [0] and matcher [x'' / x']
6.64/2.65	
6.64/2.65	H(f(f(x'))) -> H(f(g(f(x'))))
6.64/2.65	with rule H(f(f(x))) -> H(f(g(f(x))))
6.64/2.65	
6.64/2.65	Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
6.64/2.65	
6.64/2.65	
6.64/2.65	All these steps are and every following step will be a correct step w.r.t to Q.
6.64/2.65	
6.64/2.65	
6.64/2.65	
6.64/2.65	
6.64/2.65	----------------------------------------
6.64/2.65	
6.64/2.65	(10)
6.64/2.65	NO
6.64/2.69	EOF