19.86/6.06	YES
19.97/6.06	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
19.97/6.06	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
19.97/6.06	
19.97/6.06	
19.97/6.06	Termination w.r.t. Q of the given QTRS could be proven:
19.97/6.06	
19.97/6.06	(0) QTRS
19.97/6.06	(1) QTRS Reverse [EQUIVALENT, 0 ms]
19.97/6.06	(2) QTRS
19.97/6.06	(3) QTRSRRRProof [EQUIVALENT, 16 ms]
19.97/6.06	(4) QTRS
19.97/6.06	(5) DependencyPairsProof [EQUIVALENT, 9 ms]
19.97/6.06	(6) QDP
19.97/6.06	(7) DependencyGraphProof [EQUIVALENT, 0 ms]
19.97/6.06	(8) AND
19.97/6.06	    (9) QDP
19.97/6.06	        (10) UsableRulesProof [EQUIVALENT, 0 ms]
19.97/6.06	        (11) QDP
19.97/6.06	        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
19.97/6.06	        (13) YES
19.97/6.06	    (14) QDP
19.97/6.06	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
19.97/6.06	        (16) QDP
19.97/6.06	        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
19.97/6.06	        (18) YES
19.97/6.06	    (19) QDP
19.97/6.06	        (20) UsableRulesProof [EQUIVALENT, 0 ms]
19.97/6.06	        (21) QDP
19.97/6.06	        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
19.97/6.06	        (23) YES
19.97/6.06	    (24) QDP
19.97/6.06	        (25) UsableRulesProof [EQUIVALENT, 0 ms]
19.97/6.06	        (26) QDP
19.97/6.06	        (27) QDPOrderProof [EQUIVALENT, 0 ms]
19.97/6.06	        (28) QDP
19.97/6.06	        (29) PisEmptyProof [EQUIVALENT, 0 ms]
19.97/6.06	        (30) YES
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(0)
19.97/6.06	Obligation:
19.97/6.06	Q restricted rewrite system:
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   t(f(x1)) -> t(c(n(x1)))
19.97/6.06	   n(f(x1)) -> f(n(x1))
19.97/6.06	   o(f(x1)) -> f(o(x1))
19.97/6.06	   n(s(x1)) -> f(s(x1))
19.97/6.06	   o(s(x1)) -> f(s(x1))
19.97/6.06	   c(f(x1)) -> f(c(x1))
19.97/6.06	   c(n(x1)) -> n(c(x1))
19.97/6.06	   c(o(x1)) -> o(c(x1))
19.97/6.06	   c(o(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(1) QTRS Reverse (EQUIVALENT)
19.97/6.06	We applied the QTRS Reverse Processor [REVERSE].
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(2)
19.97/6.06	Obligation:
19.97/6.06	Q restricted rewrite system:
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   s(o(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(3) QTRSRRRProof (EQUIVALENT)
19.97/6.06	Used ordering:
19.97/6.06	Polynomial interpretation [POLO]:
19.97/6.06	
19.97/6.06	   POL(c(x_1)) = x_1
19.97/6.06	   POL(f(x_1)) = x_1
19.97/6.06	   POL(n(x_1)) = x_1
19.97/6.06	   POL(o(x_1)) = 1 + x_1
19.97/6.06	   POL(s(x_1)) = x_1
19.97/6.06	   POL(t(x_1)) = x_1
19.97/6.06	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
19.97/6.06	
19.97/6.06	   s(o(x1)) -> s(f(x1))
19.97/6.06	
19.97/6.06	
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(4)
19.97/6.06	Obligation:
19.97/6.06	Q restricted rewrite system:
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(5) DependencyPairsProof (EQUIVALENT)
19.97/6.06	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(6)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   F(t(x1)) -> N(c(t(x1)))
19.97/6.06	   F(n(x1)) -> N(f(x1))
19.97/6.06	   F(n(x1)) -> F(x1)
19.97/6.06	   F(o(x1)) -> O(f(x1))
19.97/6.06	   F(o(x1)) -> F(x1)
19.97/6.06	   S(n(x1)) -> S(f(x1))
19.97/6.06	   S(n(x1)) -> F(x1)
19.97/6.06	   F(c(x1)) -> F(x1)
19.97/6.06	   N(c(x1)) -> N(x1)
19.97/6.06	   O(c(x1)) -> O(x1)
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(7) DependencyGraphProof (EQUIVALENT)
19.97/6.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(8)
19.97/6.06	Complex Obligation (AND)
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(9)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   O(c(x1)) -> O(x1)
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(10) UsableRulesProof (EQUIVALENT)
19.97/6.06	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.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(11)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   O(c(x1)) -> O(x1)
19.97/6.06	
19.97/6.06	R is empty.
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(12) QDPSizeChangeProof (EQUIVALENT)
19.97/6.06	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. 
19.97/6.06	
19.97/6.06	From the DPs we obtained the following set of size-change graphs:
19.97/6.06	*O(c(x1)) -> O(x1)
19.97/6.06	The graph contains the following edges 1 > 1
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(13)
19.97/6.06	YES
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(14)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   N(c(x1)) -> N(x1)
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(15) UsableRulesProof (EQUIVALENT)
19.97/6.06	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.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(16)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   N(c(x1)) -> N(x1)
19.97/6.06	
19.97/6.06	R is empty.
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(17) QDPSizeChangeProof (EQUIVALENT)
19.97/6.06	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. 
19.97/6.06	
19.97/6.06	From the DPs we obtained the following set of size-change graphs:
19.97/6.06	*N(c(x1)) -> N(x1)
19.97/6.06	The graph contains the following edges 1 > 1
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(18)
19.97/6.06	YES
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(19)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   F(o(x1)) -> F(x1)
19.97/6.06	   F(n(x1)) -> F(x1)
19.97/6.06	   F(c(x1)) -> F(x1)
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(20) UsableRulesProof (EQUIVALENT)
19.97/6.06	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.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(21)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   F(o(x1)) -> F(x1)
19.97/6.06	   F(n(x1)) -> F(x1)
19.97/6.06	   F(c(x1)) -> F(x1)
19.97/6.06	
19.97/6.06	R is empty.
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(22) QDPSizeChangeProof (EQUIVALENT)
19.97/6.06	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. 
19.97/6.06	
19.97/6.06	From the DPs we obtained the following set of size-change graphs:
19.97/6.06	*F(o(x1)) -> F(x1)
19.97/6.06	The graph contains the following edges 1 > 1
19.97/6.06	
19.97/6.06	
19.97/6.06	*F(n(x1)) -> F(x1)
19.97/6.06	The graph contains the following edges 1 > 1
19.97/6.06	
19.97/6.06	
19.97/6.06	*F(c(x1)) -> F(x1)
19.97/6.06	The graph contains the following edges 1 > 1
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(23)
19.97/6.06	YES
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(24)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   S(n(x1)) -> S(f(x1))
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   s(n(x1)) -> s(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(25) UsableRulesProof (EQUIVALENT)
19.97/6.06	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.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(26)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	The TRS P consists of the following rules:
19.97/6.06	
19.97/6.06	   S(n(x1)) -> S(f(x1))
19.97/6.06	
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(27) QDPOrderProof (EQUIVALENT)
19.97/6.06	We use the reduction pair processor [LPAR04,JAR06].
19.97/6.06	
19.97/6.06	
19.97/6.06	The following pairs can be oriented strictly and are deleted.
19.97/6.06	
19.97/6.06	   S(n(x1)) -> S(f(x1))
19.97/6.06	The remaining pairs can at least be oriented weakly.
19.97/6.06	Used ordering:  Polynomial interpretation [POLO]:
19.97/6.06	
19.97/6.06	   POL(S(x_1)) = x_1
19.97/6.06	   POL(c(x_1)) = 0
19.97/6.06	   POL(f(x_1)) = x_1
19.97/6.06	   POL(n(x_1)) = 1 + x_1
19.97/6.06	   POL(o(x_1)) = 1
19.97/6.06	   POL(t(x_1)) = 1 + x_1
19.97/6.06	
19.97/6.06	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	
19.97/6.06	
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(28)
19.97/6.06	Obligation:
19.97/6.06	Q DP problem:
19.97/6.06	P is empty.
19.97/6.06	The TRS R consists of the following rules:
19.97/6.06	
19.97/6.06	   f(t(x1)) -> n(c(t(x1)))
19.97/6.06	   f(n(x1)) -> n(f(x1))
19.97/6.06	   f(o(x1)) -> o(f(x1))
19.97/6.06	   f(c(x1)) -> c(f(x1))
19.97/6.06	   o(c(x1)) -> c(o(x1))
19.97/6.06	   o(c(x1)) -> o(x1)
19.97/6.06	   n(c(x1)) -> c(n(x1))
19.97/6.06	
19.97/6.06	Q is empty.
19.97/6.06	We have to consider all minimal (P,Q,R)-chains.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(29) PisEmptyProof (EQUIVALENT)
19.97/6.06	The TRS P is empty. Hence, there is no (P,Q,R) chain.
19.97/6.06	----------------------------------------
19.97/6.06	
19.97/6.06	(30)
19.97/6.06	YES
20.25/9.68	EOF