3.52/1.71	YES
3.52/1.72	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.52/1.72	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.52/1.72	
3.52/1.72	
3.52/1.72	Termination w.r.t. Q of the given QTRS could be proven:
3.52/1.72	
3.52/1.72	(0) QTRS
3.52/1.72	(1) QTRSRRRProof [EQUIVALENT, 53 ms]
3.52/1.72	(2) QTRS
3.52/1.72	(3) RisEmptyProof [EQUIVALENT, 0 ms]
3.52/1.72	(4) YES
3.52/1.72	
3.52/1.72	
3.52/1.72	----------------------------------------
3.52/1.72	
3.52/1.72	(0)
3.52/1.72	Obligation:
3.52/1.72	Q restricted rewrite system:
3.52/1.72	The TRS R consists of the following rules:
3.52/1.72	
3.52/1.72	   terms(N) -> cons(recip(sqr(N)))
3.52/1.72	   sqr(0) -> 0
3.52/1.72	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.52/1.72	   dbl(0) -> 0
3.52/1.72	   dbl(s(X)) -> s(s(dbl(X)))
3.52/1.72	   add(0, X) -> X
3.52/1.72	   add(s(X), Y) -> s(add(X, Y))
3.52/1.72	   first(0, X) -> nil
3.52/1.72	   first(s(X), cons(Y)) -> cons(Y)
3.52/1.72	   half(0) -> 0
3.52/1.72	   half(s(0)) -> 0
3.52/1.72	   half(s(s(X))) -> s(half(X))
3.52/1.72	   half(dbl(X)) -> X
3.52/1.72	
3.52/1.72	The set Q consists of the following terms:
3.52/1.72	
3.52/1.72	   terms(x0)
3.52/1.72	   sqr(0)
3.52/1.72	   sqr(s(x0))
3.52/1.72	   dbl(0)
3.52/1.72	   dbl(s(x0))
3.52/1.72	   add(0, x0)
3.52/1.72	   add(s(x0), x1)
3.52/1.72	   first(0, x0)
3.52/1.72	   first(s(x0), cons(x1))
3.52/1.72	   half(0)
3.52/1.72	   half(s(0))
3.52/1.72	   half(s(s(x0)))
3.52/1.72	   half(dbl(x0))
3.52/1.72	
3.52/1.72	
3.52/1.72	----------------------------------------
3.52/1.72	
3.52/1.72	(1) QTRSRRRProof (EQUIVALENT)
3.52/1.72	Used ordering:
3.52/1.72	Quasi precedence:
3.52/1.72	terms_1 > cons_1 > s_1
3.52/1.72	terms_1 > recip_1 > s_1
3.52/1.72	terms_1 > [sqr_1, dbl_1] > [0, nil, half_1] > s_1
3.52/1.72	terms_1 > [sqr_1, dbl_1] > add_2 > s_1
3.52/1.72	first_2 > s_1
3.52/1.72	
3.52/1.72	
3.52/1.72	Status:
3.52/1.72	terms_1: multiset status
3.52/1.72	cons_1: multiset status
3.52/1.72	recip_1: multiset status
3.52/1.72	sqr_1: multiset status
3.52/1.72	0: multiset status
3.52/1.72	s_1: multiset status
3.52/1.72	add_2: [1,2]
3.52/1.72	dbl_1: multiset status
3.52/1.72	first_2: multiset status
3.52/1.72	nil: multiset status
3.52/1.72	half_1: multiset status
3.52/1.72	
3.52/1.72	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
3.52/1.72	
3.52/1.72	   terms(N) -> cons(recip(sqr(N)))
3.52/1.72	   sqr(0) -> 0
3.52/1.72	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.52/1.72	   dbl(0) -> 0
3.52/1.72	   dbl(s(X)) -> s(s(dbl(X)))
3.52/1.72	   add(0, X) -> X
3.52/1.72	   add(s(X), Y) -> s(add(X, Y))
3.52/1.72	   first(0, X) -> nil
3.52/1.72	   first(s(X), cons(Y)) -> cons(Y)
3.52/1.72	   half(0) -> 0
3.52/1.72	   half(s(0)) -> 0
3.52/1.72	   half(s(s(X))) -> s(half(X))
3.52/1.72	   half(dbl(X)) -> X
3.52/1.72	
3.52/1.72	
3.52/1.72	
3.52/1.72	
3.52/1.72	----------------------------------------
3.52/1.72	
3.52/1.72	(2)
3.52/1.72	Obligation:
3.52/1.72	Q restricted rewrite system:
3.52/1.72	R is empty.
3.52/1.72	The set Q consists of the following terms:
3.52/1.72	
3.52/1.72	   terms(x0)
3.52/1.72	   sqr(0)
3.52/1.72	   sqr(s(x0))
3.52/1.72	   dbl(0)
3.52/1.72	   dbl(s(x0))
3.52/1.72	   add(0, x0)
3.52/1.72	   add(s(x0), x1)
3.52/1.72	   first(0, x0)
3.52/1.72	   first(s(x0), cons(x1))
3.52/1.72	   half(0)
3.52/1.72	   half(s(0))
3.52/1.72	   half(s(s(x0)))
3.52/1.72	   half(dbl(x0))
3.52/1.72	
3.52/1.72	
3.52/1.72	----------------------------------------
3.52/1.72	
3.52/1.72	(3) RisEmptyProof (EQUIVALENT)
3.52/1.72	The TRS R is empty. Hence, termination is trivially proven.
3.52/1.72	----------------------------------------
3.52/1.72	
3.52/1.72	(4)
3.52/1.72	YES
3.52/1.73	EOF