3.71/1.89	MAYBE
3.97/1.89	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.97/1.89	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.97/1.89	
3.97/1.89	
3.97/1.89	Quasi decreasingness of the given CTRS could not be shown:
3.97/1.89	
3.97/1.89	(0) CTRS
3.97/1.89	(1) CTRSToQTRSProof [SOUND, 0 ms]
3.97/1.89	(2) QTRS
3.97/1.89	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.97/1.89	(4) QDP
3.97/1.89	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.97/1.89	(6) QDP
3.97/1.89	(7) NonTerminationLoopProof [COMPLETE, 0 ms]
3.97/1.89	(8) NO
3.97/1.89	
3.97/1.89	
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(0)
3.97/1.89	Obligation:
3.97/1.89	Conditional term rewrite system:
3.97/1.89	The TRS R consists of the following rules:
3.97/1.89	
3.97/1.89	   h(d) -> c(a)
3.97/1.89	   h(d) -> c(b)
3.97/1.89	   f(k(a), k(b), x) -> f(x, x, x)
3.97/1.89	
3.97/1.89	The conditional TRS C consists of the following conditional rules:
3.97/1.89	
3.97/1.89	   g(x) -> k(y) <= h(x) -> d, h(x) -> c(y)
3.97/1.89	
3.97/1.89	
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(1) CTRSToQTRSProof (SOUND)
3.97/1.89	The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(2)
3.97/1.89	Obligation:
3.97/1.89	Q restricted rewrite system:
3.97/1.89	The TRS R consists of the following rules:
3.97/1.89	
3.97/1.89	   g(x) -> U1(h(x), x)
3.97/1.89	   U1(d, x) -> U2(h(x))
3.97/1.89	   U2(c(y)) -> k(y)
3.97/1.89	   h(d) -> c(a)
3.97/1.89	   h(d) -> c(b)
3.97/1.89	   f(k(a), k(b), x) -> f(x, x, x)
3.97/1.89	
3.97/1.89	Q is empty.
3.97/1.89	
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(3) DependencyPairsProof (EQUIVALENT)
3.97/1.89	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(4)
3.97/1.89	Obligation:
3.97/1.89	Q DP problem:
3.97/1.89	The TRS P consists of the following rules:
3.97/1.89	
3.97/1.89	   G(x) -> U1^1(h(x), x)
3.97/1.89	   G(x) -> H(x)
3.97/1.89	   U1^1(d, x) -> U2^1(h(x))
3.97/1.89	   U1^1(d, x) -> H(x)
3.97/1.89	   F(k(a), k(b), x) -> F(x, x, x)
3.97/1.89	
3.97/1.89	The TRS R consists of the following rules:
3.97/1.89	
3.97/1.89	   g(x) -> U1(h(x), x)
3.97/1.89	   U1(d, x) -> U2(h(x))
3.97/1.89	   U2(c(y)) -> k(y)
3.97/1.89	   h(d) -> c(a)
3.97/1.89	   h(d) -> c(b)
3.97/1.89	   f(k(a), k(b), x) -> f(x, x, x)
3.97/1.89	
3.97/1.89	Q is empty.
3.97/1.89	We have to consider all minimal (P,Q,R)-chains.
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(5) DependencyGraphProof (EQUIVALENT)
3.97/1.89	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(6)
3.97/1.89	Obligation:
3.97/1.89	Q DP problem:
3.97/1.89	The TRS P consists of the following rules:
3.97/1.89	
3.97/1.89	   F(k(a), k(b), x) -> F(x, x, x)
3.97/1.89	
3.97/1.89	The TRS R consists of the following rules:
3.97/1.89	
3.97/1.89	   g(x) -> U1(h(x), x)
3.97/1.89	   U1(d, x) -> U2(h(x))
3.97/1.89	   U2(c(y)) -> k(y)
3.97/1.89	   h(d) -> c(a)
3.97/1.89	   h(d) -> c(b)
3.97/1.89	   f(k(a), k(b), x) -> f(x, x, x)
3.97/1.89	
3.97/1.89	Q is empty.
3.97/1.89	We have to consider all minimal (P,Q,R)-chains.
3.97/1.89	----------------------------------------
3.97/1.89	
3.97/1.89	(7) NonTerminationLoopProof (COMPLETE)
3.97/1.89	We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
3.97/1.89	Found a loop by narrowing to the left:
3.97/1.89	
3.97/1.89	s = F(U2(h(d)), U2(h(d)), x) evaluates to  t =F(x, x, x)
3.97/1.89	
3.97/1.89	Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
3.97/1.89	* Matcher: [ ]
3.97/1.89	* Semiunifier: [x / U2(h(d))]
3.97/1.89	
3.97/1.89	--------------------------------------------------------------------------------
3.97/1.89	Rewriting sequence
3.97/1.89	
3.97/1.89	F(U2(h(d)), U2(h(d)), U2(h(d))) -> F(U2(h(d)), U2(c(b)), U2(h(d)))
3.97/1.89	with rule h(d) -> c(b) at position [1,0] and matcher [ ]
3.97/1.89	
3.97/1.89	F(U2(h(d)), U2(c(b)), U2(h(d))) -> F(U2(h(d)), k(b), U2(h(d)))
3.97/1.89	with rule U2(c(y)) -> k(y) at position [1] and matcher [y / b]
3.97/1.89	
3.97/1.89	F(U2(h(d)), k(b), U2(h(d))) -> F(U2(c(a)), k(b), U2(h(d)))
3.97/1.89	with rule h(d) -> c(a) at position [0,0] and matcher [ ]
3.97/1.89	
3.97/1.89	F(U2(c(a)), k(b), U2(h(d))) -> F(k(a), k(b), U2(h(d)))
3.97/1.89	with rule U2(c(y)) -> k(y) at position [0] and matcher [y / a]
3.97/1.89	
3.97/1.89	F(k(a), k(b), U2(h(d))) -> F(U2(h(d)), U2(h(d)), U2(h(d)))
3.97/1.89	with rule F(k(a), k(b), x) -> F(x, x, x)
3.97/1.89	
3.97/1.89	Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
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3.97/1.89	
3.97/1.89	All these steps are and every following step will be a correct step w.r.t to Q.
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3.97/1.89	
3.97/1.89	(8)
3.97/1.89	NO
4.00/1.93	EOF