12.80/4.08	YES
13.12/4.14	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
13.12/4.14	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
13.12/4.14	
13.12/4.14	
13.12/4.14	Termination w.r.t. Q of the given QTRS could be proven:
13.12/4.14	
13.12/4.14	(0) QTRS
13.12/4.14	(1) QTRSRRRProof [EQUIVALENT, 13 ms]
13.12/4.14	(2) QTRS
13.12/4.14	(3) DependencyPairsProof [EQUIVALENT, 32 ms]
13.12/4.14	(4) QDP
13.12/4.14	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
13.12/4.14	(6) QDP
13.12/4.14	(7) UsableRulesProof [EQUIVALENT, 0 ms]
13.12/4.14	(8) QDP
13.12/4.14	(9) QDPSizeChangeProof [EQUIVALENT, 1 ms]
13.12/4.14	(10) YES
13.12/4.14	
13.12/4.14	
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(0)
13.12/4.14	Obligation:
13.12/4.14	Q restricted rewrite system:
13.12/4.14	The TRS R consists of the following rules:
13.12/4.14	
13.12/4.14	   1(0(x1)) -> 0(0(0(1(x1))))
13.12/4.14	   0(1(x1)) -> 1(x1)
13.12/4.14	   1(1(x1)) -> 0(0(0(0(x1))))
13.12/4.14	   0(0(x1)) -> 0(x1)
13.12/4.14	
13.12/4.14	Q is empty.
13.12/4.14	
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(1) QTRSRRRProof (EQUIVALENT)
13.12/4.14	Used ordering:
13.12/4.14	Polynomial interpretation [POLO]:
13.12/4.14	
13.12/4.14	   POL(0(x_1)) = x_1
13.12/4.14	   POL(1(x_1)) = 1 + x_1
13.12/4.14	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
13.12/4.14	
13.12/4.14	   1(1(x1)) -> 0(0(0(0(x1))))
13.12/4.14	
13.12/4.14	
13.12/4.14	
13.12/4.14	
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(2)
13.12/4.14	Obligation:
13.12/4.14	Q restricted rewrite system:
13.12/4.14	The TRS R consists of the following rules:
13.12/4.14	
13.12/4.14	   1(0(x1)) -> 0(0(0(1(x1))))
13.12/4.14	   0(1(x1)) -> 1(x1)
13.12/4.14	   0(0(x1)) -> 0(x1)
13.12/4.14	
13.12/4.14	Q is empty.
13.12/4.14	
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(3) DependencyPairsProof (EQUIVALENT)
13.12/4.14	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(4)
13.12/4.14	Obligation:
13.12/4.14	Q DP problem:
13.12/4.14	The TRS P consists of the following rules:
13.12/4.14	
13.12/4.14	   1^1(0(x1)) -> 0^1(0(0(1(x1))))
13.12/4.14	   1^1(0(x1)) -> 0^1(0(1(x1)))
13.12/4.14	   1^1(0(x1)) -> 0^1(1(x1))
13.12/4.14	   1^1(0(x1)) -> 1^1(x1)
13.12/4.14	
13.12/4.14	The TRS R consists of the following rules:
13.12/4.14	
13.12/4.14	   1(0(x1)) -> 0(0(0(1(x1))))
13.12/4.14	   0(1(x1)) -> 1(x1)
13.12/4.14	   0(0(x1)) -> 0(x1)
13.12/4.14	
13.12/4.14	Q is empty.
13.12/4.14	We have to consider all minimal (P,Q,R)-chains.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(5) DependencyGraphProof (EQUIVALENT)
13.12/4.14	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(6)
13.12/4.14	Obligation:
13.12/4.14	Q DP problem:
13.12/4.14	The TRS P consists of the following rules:
13.12/4.14	
13.12/4.14	   1^1(0(x1)) -> 1^1(x1)
13.12/4.14	
13.12/4.14	The TRS R consists of the following rules:
13.12/4.14	
13.12/4.14	   1(0(x1)) -> 0(0(0(1(x1))))
13.12/4.14	   0(1(x1)) -> 1(x1)
13.12/4.14	   0(0(x1)) -> 0(x1)
13.12/4.14	
13.12/4.14	Q is empty.
13.12/4.14	We have to consider all minimal (P,Q,R)-chains.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(7) UsableRulesProof (EQUIVALENT)
13.12/4.14	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.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(8)
13.12/4.14	Obligation:
13.12/4.14	Q DP problem:
13.12/4.14	The TRS P consists of the following rules:
13.12/4.14	
13.12/4.14	   1^1(0(x1)) -> 1^1(x1)
13.12/4.14	
13.12/4.14	R is empty.
13.12/4.14	Q is empty.
13.12/4.14	We have to consider all minimal (P,Q,R)-chains.
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(9) QDPSizeChangeProof (EQUIVALENT)
13.12/4.14	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
13.12/4.14	
13.12/4.14	From the DPs we obtained the following set of size-change graphs:
13.12/4.14	*1^1(0(x1)) -> 1^1(x1)
13.12/4.14	The graph contains the following edges 1 > 1
13.12/4.14	
13.12/4.14	
13.12/4.14	----------------------------------------
13.12/4.14	
13.12/4.14	(10)
13.12/4.14	YES
13.24/4.20	EOF