28.29/8.24	YES
28.29/8.25	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
28.29/8.25	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
28.29/8.25	
28.29/8.25	
28.29/8.25	Termination w.r.t. Q of the given QTRS could be proven:
28.29/8.25	
28.29/8.25	(0) QTRS
28.29/8.25	(1) QTRSRRRProof [EQUIVALENT, 68 ms]
28.29/8.25	(2) QTRS
28.29/8.25	(3) DependencyPairsProof [EQUIVALENT, 35 ms]
28.29/8.25	(4) QDP
28.29/8.25	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
28.29/8.25	(6) AND
28.29/8.25	    (7) QDP
28.29/8.25	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
28.29/8.25	        (9) QDP
28.29/8.25	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
28.29/8.25	        (11) YES
28.29/8.25	    (12) QDP
28.29/8.25	        (13) MNOCProof [EQUIVALENT, 0 ms]
28.29/8.25	        (14) QDP
28.29/8.25	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
28.29/8.25	        (16) QDP
28.29/8.25	        (17) QReductionProof [EQUIVALENT, 0 ms]
28.29/8.25	        (18) QDP
28.29/8.25	        (19) QDPOrderProof [EQUIVALENT, 34 ms]
28.29/8.25	        (20) QDP
28.29/8.25	        (21) PisEmptyProof [EQUIVALENT, 0 ms]
28.29/8.25	        (22) YES
28.29/8.25	
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(0)
28.29/8.25	Obligation:
28.29/8.25	Q restricted rewrite system:
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(0(x1)) -> p(p(s(s(0(s(s(s(s(x1)))))))))
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(0(x1)) -> p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	Q is empty.
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(1) QTRSRRRProof (EQUIVALENT)
28.29/8.25	Used ordering:
28.29/8.25	Polynomial interpretation [POLO]:
28.29/8.25	
28.29/8.25	   POL(0(x_1)) = x_1
28.29/8.25	   POL(p(x_1)) = x_1
28.29/8.25	   POL(q(x_1)) = 1 + x_1
28.29/8.25	   POL(r(x_1)) = 1 + x_1
28.29/8.25	   POL(s(x_1)) = x_1
28.29/8.25	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
28.29/8.25	
28.29/8.25	   q(0(x1)) -> p(p(s(s(0(s(s(s(s(x1)))))))))
28.29/8.25	   r(0(x1)) -> p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
28.29/8.25	
28.29/8.25	
28.29/8.25	
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(2)
28.29/8.25	Obligation:
28.29/8.25	Q restricted rewrite system:
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	Q is empty.
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(3) DependencyPairsProof (EQUIVALENT)
28.29/8.25	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(4)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> P(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   Q(s(x1)) -> P(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   Q(s(x1)) -> P(p(s(s(x1))))
28.29/8.25	   Q(s(x1)) -> P(s(s(x1)))
28.29/8.25	   R(s(x1)) -> P(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   R(s(x1)) -> P(s(s(q(p(s(p(s(x1))))))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	   R(s(x1)) -> P(s(p(s(x1))))
28.29/8.25	   R(s(x1)) -> P(s(x1))
28.29/8.25	   P(p(s(x1))) -> P(x1)
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	Q is empty.
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(5) DependencyGraphProof (EQUIVALENT)
28.29/8.25	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(6)
28.29/8.25	Complex Obligation (AND)
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(7)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   P(p(s(x1))) -> P(x1)
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	Q is empty.
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(8) UsableRulesProof (EQUIVALENT)
28.29/8.25	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.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(9)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   P(p(s(x1))) -> P(x1)
28.29/8.25	
28.29/8.25	R is empty.
28.29/8.25	Q is empty.
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(10) QDPSizeChangeProof (EQUIVALENT)
28.29/8.25	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. 
28.29/8.25	
28.29/8.25	From the DPs we obtained the following set of size-change graphs:
28.29/8.25	*P(p(s(x1))) -> P(x1)
28.29/8.25	The graph contains the following edges 1 > 1
28.29/8.25	
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(11)
28.29/8.25	YES
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(12)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	Q is empty.
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(13) MNOCProof (EQUIVALENT)
28.29/8.25	We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(14)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
28.29/8.25	   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
28.29/8.25	   p(p(s(x1))) -> p(x1)
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	The set Q consists of the following terms:
28.29/8.25	
28.29/8.25	   q(s(x0))
28.29/8.25	   r(s(x0))
28.29/8.25	   p(s(x0))
28.29/8.25	   p(0(x0))
28.29/8.25	
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(15) UsableRulesProof (EQUIVALENT)
28.29/8.25	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(16)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	The set Q consists of the following terms:
28.29/8.25	
28.29/8.25	   q(s(x0))
28.29/8.25	   r(s(x0))
28.29/8.25	   p(s(x0))
28.29/8.25	   p(0(x0))
28.29/8.25	
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(17) QReductionProof (EQUIVALENT)
28.29/8.25	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
28.29/8.25	
28.29/8.25	   q(s(x0))
28.29/8.25	   r(s(x0))
28.29/8.25	
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(18)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	The TRS P consists of the following rules:
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	The set Q consists of the following terms:
28.29/8.25	
28.29/8.25	   p(s(x0))
28.29/8.25	   p(0(x0))
28.29/8.25	
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(19) QDPOrderProof (EQUIVALENT)
28.29/8.25	We use the reduction pair processor [LPAR04,JAR06].
28.29/8.25	
28.29/8.25	
28.29/8.25	The following pairs can be oriented strictly and are deleted.
28.29/8.25	
28.29/8.25	   Q(s(x1)) -> R(p(p(s(s(x1)))))
28.29/8.25	   R(s(x1)) -> Q(p(s(p(s(x1)))))
28.29/8.25	The remaining pairs can at least be oriented weakly.
28.29/8.25	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
28.29/8.25	
28.29/8.25	POL( Q_1(x_1) ) = max{0, 2x_1 - 2}
28.29/8.25	POL( R_1(x_1) ) = max{0, 2x_1 - 2}
28.29/8.25	POL( p_1(x_1) ) = max{0, x_1 - 2}
28.29/8.25	POL( s_1(x_1) ) = x_1 + 2
28.29/8.25	POL( 0_1(x_1) ) = max{0, -2}
28.29/8.25	
28.29/8.25	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
28.29/8.25	
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(20)
28.29/8.25	Obligation:
28.29/8.25	Q DP problem:
28.29/8.25	P is empty.
28.29/8.25	The TRS R consists of the following rules:
28.29/8.25	
28.29/8.25	   p(s(x1)) -> x1
28.29/8.25	   p(0(x1)) -> 0(s(s(s(x1))))
28.29/8.25	
28.29/8.25	The set Q consists of the following terms:
28.29/8.25	
28.29/8.25	   p(s(x0))
28.29/8.25	   p(0(x0))
28.29/8.25	
28.29/8.25	We have to consider all minimal (P,Q,R)-chains.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(21) PisEmptyProof (EQUIVALENT)
28.29/8.25	The TRS P is empty. Hence, there is no (P,Q,R) chain.
28.29/8.25	----------------------------------------
28.29/8.25	
28.29/8.25	(22)
28.29/8.25	YES
28.29/8.34	EOF