3.26/1.58	YES
3.26/1.59	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.26/1.59	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.26/1.59	
3.26/1.59	
3.26/1.59	Termination w.r.t. Q of the given QTRS could be proven:
3.26/1.59	
3.26/1.59	(0) QTRS
3.26/1.59	(1) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
3.26/1.59	(2) YES
3.26/1.59	
3.26/1.59	
3.26/1.59	----------------------------------------
3.26/1.59	
3.26/1.59	(0)
3.26/1.59	Obligation:
3.26/1.59	Q restricted rewrite system:
3.26/1.59	The TRS R consists of the following rules:
3.26/1.59	
3.26/1.59	   active(g(X)) -> mark(h(X))
3.26/1.59	   active(c) -> mark(d)
3.26/1.59	   active(h(d)) -> mark(g(c))
3.26/1.59	   mark(g(X)) -> active(g(X))
3.26/1.59	   mark(h(X)) -> active(h(X))
3.26/1.59	   mark(c) -> active(c)
3.26/1.59	   mark(d) -> active(d)
3.26/1.59	   g(mark(X)) -> g(X)
3.26/1.59	   g(active(X)) -> g(X)
3.26/1.59	   h(mark(X)) -> h(X)
3.26/1.59	   h(active(X)) -> h(X)
3.26/1.59	
3.26/1.59	The set Q consists of the following terms:
3.26/1.59	
3.26/1.59	   active(g(x0))
3.26/1.59	   active(c)
3.26/1.59	   active(h(d))
3.26/1.59	   mark(g(x0))
3.26/1.59	   mark(h(x0))
3.26/1.59	   mark(c)
3.26/1.59	   mark(d)
3.26/1.59	   g(mark(x0))
3.26/1.59	   g(active(x0))
3.26/1.59	   h(mark(x0))
3.26/1.59	   h(active(x0))
3.26/1.59	
3.26/1.59	
3.26/1.59	----------------------------------------
3.26/1.59	
3.26/1.59	(1) RFCMatchBoundsTRSProof (EQUIVALENT)
3.26/1.59	Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 4. This implies Q-termination of R.
3.26/1.59	The following rules were used to construct the certificate:
3.26/1.59	
3.26/1.59	   active(g(X)) -> mark(h(X))
3.26/1.59	   active(c) -> mark(d)
3.26/1.59	   active(h(d)) -> mark(g(c))
3.26/1.59	   mark(g(X)) -> active(g(X))
3.26/1.59	   mark(h(X)) -> active(h(X))
3.26/1.59	   mark(c) -> active(c)
3.26/1.59	   mark(d) -> active(d)
3.26/1.59	   g(mark(X)) -> g(X)
3.26/1.59	   g(active(X)) -> g(X)
3.26/1.59	   h(mark(X)) -> h(X)
3.26/1.59	   h(active(X)) -> h(X)
3.26/1.59	
3.26/1.59	The certificate found is represented by the following graph.
3.26/1.59	The certificate consists of the following enumerated nodes:
3.26/1.59	1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
3.26/1.59	
3.26/1.59	Node 1 is start node and node 2 is final node.
3.26/1.59	
3.26/1.59	Those nodes are connected through the following edges:
3.26/1.59	
3.26/1.59	* 1 to 3 labelled mark_1(0), active_1(0), active_1(1), mark_1(1)* 1 to 4 labelled mark_1(0)* 1 to 6 labelled active_1(0)* 1 to 5 labelled active_1(0)* 1 to 2 labelled g_1(0), h_1(0), g_1(1), h_1(1)* 1 to 7 labelled active_1(1), mark_1(1)* 1 to 8 labelled active_1(2)* 1 to 9 labelled mark_1(2)* 1 to 10 labelled mark_1(3), active_1(3)* 1 to 11 labelled active_1(4)* 2 to 2 labelled #_1(0)* 3 to 2 labelled h_1(0), d(0), h_1(1)* 4 to 5 labelled g_1(0)* 5 to 2 labelled c(0), c(1)* 6 to 2 labelled g_1(0), g_1(1)* 7 to 2 labelled h_1(1), d(1)* 7 to 5 labelled g_1(1)* 8 to 2 labelled h_1(2), d(2), h_1(1)* 8 to 5 labelled g_1(2)* 9 to 5 labelled h_1(2)* 10 to 5 labelled h_1(3)* 11 to 5 labelled h_1(4)
3.26/1.59	
3.26/1.59	
3.26/1.59	----------------------------------------
3.26/1.59	
3.26/1.59	(2)
3.26/1.59	YES
3.26/1.63	EOF