4.68/2.15	MAYBE
4.68/2.16	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
4.68/2.16	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.68/2.16	
4.68/2.16	
4.68/2.16	Quasi decreasingness of the given CTRS could not be shown:
4.68/2.16	
4.68/2.16	(0) CTRS
4.68/2.16	(1) CTRSToQTRSProof [SOUND, 0 ms]
4.68/2.16	(2) QTRS
4.68/2.16	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
4.68/2.16	(4) QDP
4.68/2.16	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
4.68/2.16	(6) QDP
4.68/2.16	(7) UsableRulesProof [EQUIVALENT, 0 ms]
4.68/2.16	(8) QDP
4.68/2.16	(9) NonTerminationLoopProof [COMPLETE, 0 ms]
4.68/2.16	(10) NO
4.68/2.16	
4.68/2.16	
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(0)
4.68/2.16	Obligation:
4.68/2.16	Conditional term rewrite system:
4.68/2.16	The TRS R consists of the following rules:
4.68/2.16	
4.68/2.16	   a -> b
4.68/2.16	   f(a) -> b
4.68/2.16	
4.68/2.16	The conditional TRS C consists of the following conditional rules:
4.68/2.16	
4.68/2.16	   g(x) -> g(a) <= f(x) -> x
4.68/2.16	
4.68/2.16	
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(1) CTRSToQTRSProof (SOUND)
4.68/2.16	The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(2)
4.68/2.16	Obligation:
4.68/2.16	Q restricted rewrite system:
4.68/2.16	The TRS R consists of the following rules:
4.68/2.16	
4.68/2.16	   g(x) -> U1(f(x))
4.68/2.16	   U1(x) -> g(a)
4.68/2.16	   a -> b
4.68/2.16	   f(a) -> b
4.68/2.16	
4.68/2.16	Q is empty.
4.68/2.16	
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(3) DependencyPairsProof (EQUIVALENT)
4.68/2.16	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(4)
4.68/2.16	Obligation:
4.68/2.16	Q DP problem:
4.68/2.16	The TRS P consists of the following rules:
4.68/2.16	
4.68/2.16	   G(x) -> U1^1(f(x))
4.68/2.16	   G(x) -> F(x)
4.68/2.16	   U1^1(x) -> G(a)
4.68/2.16	   U1^1(x) -> A
4.68/2.16	
4.68/2.16	The TRS R consists of the following rules:
4.68/2.16	
4.68/2.16	   g(x) -> U1(f(x))
4.68/2.16	   U1(x) -> g(a)
4.68/2.16	   a -> b
4.68/2.16	   f(a) -> b
4.68/2.16	
4.68/2.16	Q is empty.
4.68/2.16	We have to consider all minimal (P,Q,R)-chains.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(5) DependencyGraphProof (EQUIVALENT)
4.68/2.16	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(6)
4.68/2.16	Obligation:
4.68/2.16	Q DP problem:
4.68/2.16	The TRS P consists of the following rules:
4.68/2.16	
4.68/2.16	   U1^1(x) -> G(a)
4.68/2.16	   G(x) -> U1^1(f(x))
4.68/2.16	
4.68/2.16	The TRS R consists of the following rules:
4.68/2.16	
4.68/2.16	   g(x) -> U1(f(x))
4.68/2.16	   U1(x) -> g(a)
4.68/2.16	   a -> b
4.68/2.16	   f(a) -> b
4.68/2.16	
4.68/2.16	Q is empty.
4.68/2.16	We have to consider all minimal (P,Q,R)-chains.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(7) UsableRulesProof (EQUIVALENT)
4.68/2.16	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.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(8)
4.68/2.16	Obligation:
4.68/2.16	Q DP problem:
4.68/2.16	The TRS P consists of the following rules:
4.68/2.16	
4.68/2.16	   U1^1(x) -> G(a)
4.68/2.16	   G(x) -> U1^1(f(x))
4.68/2.16	
4.68/2.16	The TRS R consists of the following rules:
4.68/2.16	
4.68/2.16	   f(a) -> b
4.68/2.16	   a -> b
4.68/2.16	
4.68/2.16	Q is empty.
4.68/2.16	We have to consider all minimal (P,Q,R)-chains.
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(9) NonTerminationLoopProof (COMPLETE)
4.68/2.16	We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
4.68/2.16	Found a loop by narrowing to the left:
4.68/2.16	
4.68/2.16	s = G(x') evaluates to  t =G(a)
4.68/2.16	
4.68/2.16	Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
4.68/2.16	* Matcher: [x' / a]
4.68/2.16	* Semiunifier: [ ]
4.68/2.16	
4.68/2.16	--------------------------------------------------------------------------------
4.68/2.16	Rewriting sequence
4.68/2.16	
4.68/2.16	G(x') -> U1^1(f(x'))
4.68/2.16	with rule G(x'') -> U1^1(f(x'')) at position [] and matcher [x'' / x']
4.68/2.16	
4.68/2.16	U1^1(f(x')) -> G(a)
4.68/2.16	with rule U1^1(x) -> G(a)
4.68/2.16	
4.68/2.16	Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
4.68/2.16	
4.68/2.16	
4.68/2.16	All these steps are and every following step will be a correct step w.r.t to Q.
4.68/2.16	
4.68/2.16	
4.68/2.16	
4.68/2.16	
4.68/2.16	----------------------------------------
4.68/2.16	
4.68/2.16	(10)
4.68/2.16	NO
4.86/2.23	EOF