3.50/1.80	YES
3.68/1.81	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.68/1.81	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.68/1.81	
3.68/1.81	
3.68/1.81	Termination w.r.t. Q of the given QTRS could be proven:
3.68/1.81	
3.68/1.81	(0) QTRS
3.68/1.81	(1) QTRSToCSRProof [SOUND, 0 ms]
3.68/1.81	(2) CSR
3.68/1.81	(3) CSRInnermostProof [EQUIVALENT, 0 ms]
3.68/1.81	(4) CSR
3.68/1.81	(5) CSDependencyPairsProof [EQUIVALENT, 0 ms]
3.68/1.81	(6) QCSDP
3.68/1.81	(7) QCSDependencyGraphProof [EQUIVALENT, 1 ms]
3.68/1.81	(8) AND
3.68/1.81	    (9) QCSDP
3.68/1.81	        (10) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.68/1.81	        (11) QCSDP
3.68/1.81	        (12) PIsEmptyProof [EQUIVALENT, 0 ms]
3.68/1.81	        (13) YES
3.68/1.81	    (14) QCSDP
3.68/1.81	        (15) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.68/1.81	        (16) QCSDP
3.68/1.81	        (17) PIsEmptyProof [EQUIVALENT, 0 ms]
3.68/1.81	        (18) YES
3.68/1.81	    (19) QCSDP
3.68/1.81	        (20) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.68/1.81	        (21) QCSDP
3.68/1.81	        (22) PIsEmptyProof [EQUIVALENT, 0 ms]
3.68/1.81	        (23) YES
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(0)
3.68/1.81	Obligation:
3.68/1.81	Q restricted rewrite system:
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
3.68/1.81	   active(sqr(0)) -> mark(0)
3.68/1.81	   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
3.68/1.81	   active(dbl(0)) -> mark(0)
3.68/1.81	   active(dbl(s(X))) -> mark(s(s(dbl(X))))
3.68/1.81	   active(add(0, X)) -> mark(X)
3.68/1.81	   active(add(s(X), Y)) -> mark(s(add(X, Y)))
3.68/1.81	   active(first(0, X)) -> mark(nil)
3.68/1.81	   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
3.68/1.81	   active(terms(X)) -> terms(active(X))
3.68/1.81	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.68/1.81	   active(recip(X)) -> recip(active(X))
3.68/1.81	   active(sqr(X)) -> sqr(active(X))
3.68/1.81	   active(s(X)) -> s(active(X))
3.68/1.81	   active(add(X1, X2)) -> add(active(X1), X2)
3.68/1.81	   active(add(X1, X2)) -> add(X1, active(X2))
3.68/1.81	   active(dbl(X)) -> dbl(active(X))
3.68/1.81	   active(first(X1, X2)) -> first(active(X1), X2)
3.68/1.81	   active(first(X1, X2)) -> first(X1, active(X2))
3.68/1.81	   terms(mark(X)) -> mark(terms(X))
3.68/1.81	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.68/1.81	   recip(mark(X)) -> mark(recip(X))
3.68/1.81	   sqr(mark(X)) -> mark(sqr(X))
3.68/1.81	   s(mark(X)) -> mark(s(X))
3.68/1.81	   add(mark(X1), X2) -> mark(add(X1, X2))
3.68/1.81	   add(X1, mark(X2)) -> mark(add(X1, X2))
3.68/1.81	   dbl(mark(X)) -> mark(dbl(X))
3.68/1.81	   first(mark(X1), X2) -> mark(first(X1, X2))
3.68/1.81	   first(X1, mark(X2)) -> mark(first(X1, X2))
3.68/1.81	   proper(terms(X)) -> terms(proper(X))
3.68/1.81	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.68/1.81	   proper(recip(X)) -> recip(proper(X))
3.68/1.81	   proper(sqr(X)) -> sqr(proper(X))
3.68/1.81	   proper(s(X)) -> s(proper(X))
3.68/1.81	   proper(0) -> ok(0)
3.68/1.81	   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
3.68/1.81	   proper(dbl(X)) -> dbl(proper(X))
3.68/1.81	   proper(first(X1, X2)) -> first(proper(X1), proper(X2))
3.68/1.81	   proper(nil) -> ok(nil)
3.68/1.81	   terms(ok(X)) -> ok(terms(X))
3.68/1.81	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.68/1.81	   recip(ok(X)) -> ok(recip(X))
3.68/1.81	   sqr(ok(X)) -> ok(sqr(X))
3.68/1.81	   s(ok(X)) -> ok(s(X))
3.68/1.81	   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
3.68/1.81	   dbl(ok(X)) -> ok(dbl(X))
3.68/1.81	   first(ok(X1), ok(X2)) -> ok(first(X1, X2))
3.68/1.81	   top(mark(X)) -> top(proper(X))
3.68/1.81	   top(ok(X)) -> top(active(X))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   active(terms(x0))
3.68/1.81	   active(cons(x0, x1))
3.68/1.81	   active(recip(x0))
3.68/1.81	   active(sqr(x0))
3.68/1.81	   active(s(x0))
3.68/1.81	   active(add(x0, x1))
3.68/1.81	   active(dbl(x0))
3.68/1.81	   active(first(x0, x1))
3.68/1.81	   terms(mark(x0))
3.68/1.81	   cons(mark(x0), x1)
3.68/1.81	   recip(mark(x0))
3.68/1.81	   sqr(mark(x0))
3.68/1.81	   s(mark(x0))
3.68/1.81	   add(mark(x0), x1)
3.68/1.81	   add(x0, mark(x1))
3.68/1.81	   dbl(mark(x0))
3.68/1.81	   first(mark(x0), x1)
3.68/1.81	   first(x0, mark(x1))
3.68/1.81	   proper(terms(x0))
3.68/1.81	   proper(cons(x0, x1))
3.68/1.81	   proper(recip(x0))
3.68/1.81	   proper(sqr(x0))
3.68/1.81	   proper(s(x0))
3.68/1.81	   proper(0)
3.68/1.81	   proper(add(x0, x1))
3.68/1.81	   proper(dbl(x0))
3.68/1.81	   proper(first(x0, x1))
3.68/1.81	   proper(nil)
3.68/1.81	   terms(ok(x0))
3.68/1.81	   cons(ok(x0), ok(x1))
3.68/1.81	   recip(ok(x0))
3.68/1.81	   sqr(ok(x0))
3.68/1.81	   s(ok(x0))
3.68/1.81	   add(ok(x0), ok(x1))
3.68/1.81	   dbl(ok(x0))
3.68/1.81	   first(ok(x0), ok(x1))
3.68/1.81	   top(mark(x0))
3.68/1.81	   top(ok(x0))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(1) QTRSToCSRProof (SOUND)
3.68/1.81	The following Q TRS is given: Q restricted rewrite system:
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
3.68/1.81	   active(sqr(0)) -> mark(0)
3.68/1.81	   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
3.68/1.81	   active(dbl(0)) -> mark(0)
3.68/1.81	   active(dbl(s(X))) -> mark(s(s(dbl(X))))
3.68/1.81	   active(add(0, X)) -> mark(X)
3.68/1.81	   active(add(s(X), Y)) -> mark(s(add(X, Y)))
3.68/1.81	   active(first(0, X)) -> mark(nil)
3.68/1.81	   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
3.68/1.81	   active(terms(X)) -> terms(active(X))
3.68/1.81	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.68/1.81	   active(recip(X)) -> recip(active(X))
3.68/1.81	   active(sqr(X)) -> sqr(active(X))
3.68/1.81	   active(s(X)) -> s(active(X))
3.68/1.81	   active(add(X1, X2)) -> add(active(X1), X2)
3.68/1.81	   active(add(X1, X2)) -> add(X1, active(X2))
3.68/1.81	   active(dbl(X)) -> dbl(active(X))
3.68/1.81	   active(first(X1, X2)) -> first(active(X1), X2)
3.68/1.81	   active(first(X1, X2)) -> first(X1, active(X2))
3.68/1.81	   terms(mark(X)) -> mark(terms(X))
3.68/1.81	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.68/1.81	   recip(mark(X)) -> mark(recip(X))
3.68/1.81	   sqr(mark(X)) -> mark(sqr(X))
3.68/1.81	   s(mark(X)) -> mark(s(X))
3.68/1.81	   add(mark(X1), X2) -> mark(add(X1, X2))
3.68/1.81	   add(X1, mark(X2)) -> mark(add(X1, X2))
3.68/1.81	   dbl(mark(X)) -> mark(dbl(X))
3.68/1.81	   first(mark(X1), X2) -> mark(first(X1, X2))
3.68/1.81	   first(X1, mark(X2)) -> mark(first(X1, X2))
3.68/1.81	   proper(terms(X)) -> terms(proper(X))
3.68/1.81	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.68/1.81	   proper(recip(X)) -> recip(proper(X))
3.68/1.81	   proper(sqr(X)) -> sqr(proper(X))
3.68/1.81	   proper(s(X)) -> s(proper(X))
3.68/1.81	   proper(0) -> ok(0)
3.68/1.81	   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
3.68/1.81	   proper(dbl(X)) -> dbl(proper(X))
3.68/1.81	   proper(first(X1, X2)) -> first(proper(X1), proper(X2))
3.68/1.81	   proper(nil) -> ok(nil)
3.68/1.81	   terms(ok(X)) -> ok(terms(X))
3.68/1.81	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.68/1.81	   recip(ok(X)) -> ok(recip(X))
3.68/1.81	   sqr(ok(X)) -> ok(sqr(X))
3.68/1.81	   s(ok(X)) -> ok(s(X))
3.68/1.81	   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
3.68/1.81	   dbl(ok(X)) -> ok(dbl(X))
3.68/1.81	   first(ok(X1), ok(X2)) -> ok(first(X1, X2))
3.68/1.81	   top(mark(X)) -> top(proper(X))
3.68/1.81	   top(ok(X)) -> top(active(X))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   active(terms(x0))
3.68/1.81	   active(cons(x0, x1))
3.68/1.81	   active(recip(x0))
3.68/1.81	   active(sqr(x0))
3.68/1.81	   active(s(x0))
3.68/1.81	   active(add(x0, x1))
3.68/1.81	   active(dbl(x0))
3.68/1.81	   active(first(x0, x1))
3.68/1.81	   terms(mark(x0))
3.68/1.81	   cons(mark(x0), x1)
3.68/1.81	   recip(mark(x0))
3.68/1.81	   sqr(mark(x0))
3.68/1.81	   s(mark(x0))
3.68/1.81	   add(mark(x0), x1)
3.68/1.81	   add(x0, mark(x1))
3.68/1.81	   dbl(mark(x0))
3.68/1.81	   first(mark(x0), x1)
3.68/1.81	   first(x0, mark(x1))
3.68/1.81	   proper(terms(x0))
3.68/1.81	   proper(cons(x0, x1))
3.68/1.81	   proper(recip(x0))
3.68/1.81	   proper(sqr(x0))
3.68/1.81	   proper(s(x0))
3.68/1.81	   proper(0)
3.68/1.81	   proper(add(x0, x1))
3.68/1.81	   proper(dbl(x0))
3.68/1.81	   proper(first(x0, x1))
3.68/1.81	   proper(nil)
3.68/1.81	   terms(ok(x0))
3.68/1.81	   cons(ok(x0), ok(x1))
3.68/1.81	   recip(ok(x0))
3.68/1.81	   sqr(ok(x0))
3.68/1.81	   s(ok(x0))
3.68/1.81	   add(ok(x0), ok(x1))
3.68/1.81	   dbl(ok(x0))
3.68/1.81	   first(ok(x0), ok(x1))
3.68/1.81	   top(mark(x0))
3.68/1.81	   top(ok(x0))
3.68/1.81	
3.68/1.81	Special symbols used for the transformation (see [GM04]):
3.68/1.81	top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
3.68/1.81	The replacement map contains the following entries:
3.68/1.81	
3.68/1.81	terms: {1}
3.68/1.81	cons: {1}
3.68/1.81	recip: {1}
3.68/1.81	sqr: {1}
3.68/1.81	s: {1}
3.68/1.81	0: empty set
3.68/1.81	add: {1, 2}
3.68/1.81	dbl: {1}
3.68/1.81	first: {1, 2}
3.68/1.81	nil: empty set
3.68/1.81	The QTRS contained just a subset of rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is sound, but not necessarily complete.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(2)
3.68/1.81	Obligation:
3.68/1.81	Context-sensitive rewrite system:
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The replacement map contains the following entries:
3.68/1.81	
3.68/1.81	terms: {1}
3.68/1.81	cons: {1}
3.68/1.81	recip: {1}
3.68/1.81	sqr: {1}
3.68/1.81	s: {1}
3.68/1.81	0: empty set
3.68/1.81	add: {1, 2}
3.68/1.81	dbl: {1}
3.68/1.81	first: {1, 2}
3.68/1.81	nil: empty set
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(3) CSRInnermostProof (EQUIVALENT)
3.68/1.81	The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(4)
3.68/1.81	Obligation:
3.68/1.81	Context-sensitive rewrite system:
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The replacement map contains the following entries:
3.68/1.81	
3.68/1.81	terms: {1}
3.68/1.81	cons: {1}
3.68/1.81	recip: {1}
3.68/1.81	sqr: {1}
3.68/1.81	s: {1}
3.68/1.81	0: empty set
3.68/1.81	add: {1, 2}
3.68/1.81	dbl: {1}
3.68/1.81	first: {1, 2}
3.68/1.81	nil: empty set
3.68/1.81	
3.68/1.81	
3.68/1.81	Innermost Strategy.
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(5) CSDependencyPairsProof (EQUIVALENT)
3.68/1.81	Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(6)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, SQR_1, TERMS_1, ADD_2, DBL_1} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The ordinary context-sensitive dependency pairs DP_o are:
3.68/1.81	   TERMS(N) -> SQR(N)
3.68/1.81	   SQR(s(X)) -> ADD(sqr(X), dbl(X))
3.68/1.81	   SQR(s(X)) -> SQR(X)
3.68/1.81	   SQR(s(X)) -> DBL(X)
3.68/1.81	   DBL(s(X)) -> DBL(X)
3.68/1.81	   ADD(s(X), Y) -> ADD(X, Y)
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(7) QCSDependencyGraphProof (EQUIVALENT)
3.68/1.81	The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 3 SCCs with 3 less nodes.
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(8)
3.68/1.81	Complex Obligation (AND)
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(9)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, DBL_1} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	
3.68/1.81	   DBL(s(X)) -> DBL(X)
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(10) QCSDPSubtermProof (EQUIVALENT)
3.68/1.81	We use the subterm processor [DA_EMMES].
3.68/1.81	
3.68/1.81	
3.68/1.81	The following pairs can be oriented strictly and are deleted.
3.68/1.81	
3.68/1.81	   DBL(s(X)) -> DBL(X)
3.68/1.81	The remaining pairs can at least be oriented weakly.
3.68/1.81	none
3.68/1.81	Used ordering:  Combined order from the following AFS and order.
3.68/1.81	DBL(x1)  =  x1
3.68/1.81	
3.68/1.81	
3.68/1.81	Subterm Order
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(11)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	none
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(12) PIsEmptyProof (EQUIVALENT)
3.68/1.81	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(13)
3.68/1.81	YES
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(14)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, ADD_2} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	
3.68/1.81	   ADD(s(X), Y) -> ADD(X, Y)
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(15) QCSDPSubtermProof (EQUIVALENT)
3.68/1.81	We use the subterm processor [DA_EMMES].
3.68/1.81	
3.68/1.81	
3.68/1.81	The following pairs can be oriented strictly and are deleted.
3.68/1.81	
3.68/1.81	   ADD(s(X), Y) -> ADD(X, Y)
3.68/1.81	The remaining pairs can at least be oriented weakly.
3.68/1.81	none
3.68/1.81	Used ordering:  Combined order from the following AFS and order.
3.68/1.81	ADD(x1, x2)  =  x1
3.68/1.81	
3.68/1.81	
3.68/1.81	Subterm Order
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(16)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	none
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(17) PIsEmptyProof (EQUIVALENT)
3.68/1.81	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(18)
3.68/1.81	YES
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(19)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, SQR_1} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	
3.68/1.81	   SQR(s(X)) -> SQR(X)
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(20) QCSDPSubtermProof (EQUIVALENT)
3.68/1.81	We use the subterm processor [DA_EMMES].
3.68/1.81	
3.68/1.81	
3.68/1.81	The following pairs can be oriented strictly and are deleted.
3.68/1.81	
3.68/1.81	   SQR(s(X)) -> SQR(X)
3.68/1.81	The remaining pairs can at least be oriented weakly.
3.68/1.81	none
3.68/1.81	Used ordering:  Combined order from the following AFS and order.
3.68/1.81	SQR(x1)  =  x1
3.68/1.81	
3.68/1.81	
3.68/1.81	Subterm Order
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(21)
3.68/1.81	Obligation:
3.68/1.81	Q-restricted context-sensitive dependency pair problem:
3.68/1.81	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
3.68/1.81	For all symbols f in {cons_2} we have mu(f) = {1}.
3.68/1.81	
3.68/1.81	The TRS P consists of the following rules:
3.68/1.81	none
3.68/1.81	
3.68/1.81	The TRS R consists of the following rules:
3.68/1.81	
3.68/1.81	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.68/1.81	   sqr(0) -> 0
3.68/1.81	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.68/1.81	   dbl(0) -> 0
3.68/1.81	   dbl(s(X)) -> s(s(dbl(X)))
3.68/1.81	   add(0, X) -> X
3.68/1.81	   add(s(X), Y) -> s(add(X, Y))
3.68/1.81	   first(0, X) -> nil
3.68/1.81	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.68/1.81	
3.68/1.81	The set Q consists of the following terms:
3.68/1.81	
3.68/1.81	   terms(x0)
3.68/1.81	   sqr(0)
3.68/1.81	   sqr(s(x0))
3.68/1.81	   dbl(0)
3.68/1.81	   dbl(s(x0))
3.68/1.81	   add(0, x0)
3.68/1.81	   add(s(x0), x1)
3.68/1.81	   first(0, x0)
3.68/1.81	   first(s(x0), cons(x1, x2))
3.68/1.81	
3.68/1.81	
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(22) PIsEmptyProof (EQUIVALENT)
3.68/1.81	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.68/1.81	----------------------------------------
3.68/1.81	
3.68/1.81	(23)
3.68/1.81	YES
3.68/1.84	EOF