3.32/1.59	YES
3.32/1.60	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.32/1.60	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.32/1.60	
3.32/1.60	
3.32/1.60	Termination of the given CSR could be proven:
3.32/1.60	
3.32/1.60	(0) CSR
3.32/1.60	(1) CSDependencyPairsProof [EQUIVALENT, 0 ms]
3.32/1.60	(2) QCSDP
3.32/1.60	(3) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
3.32/1.60	(4) AND
3.32/1.60	    (5) QCSDP
3.32/1.60	        (6) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.32/1.60	        (7) QCSDP
3.32/1.60	        (8) PIsEmptyProof [EQUIVALENT, 0 ms]
3.32/1.60	        (9) YES
3.32/1.60	    (10) QCSDP
3.32/1.60	        (11) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.32/1.60	        (12) QCSDP
3.32/1.60	        (13) PIsEmptyProof [EQUIVALENT, 0 ms]
3.32/1.60	        (14) YES
3.32/1.60	    (15) QCSDP
3.32/1.60	        (16) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.32/1.60	        (17) QCSDP
3.32/1.60	        (18) PIsEmptyProof [EQUIVALENT, 0 ms]
3.32/1.60	        (19) YES
3.32/1.60	    (20) QCSDP
3.32/1.60	        (21) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.32/1.60	        (22) QCSDP
3.32/1.60	        (23) PIsEmptyProof [EQUIVALENT, 0 ms]
3.32/1.60	        (24) YES
3.32/1.60	
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(0)
3.32/1.60	Obligation:
3.32/1.60	Context-sensitive rewrite system:
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	The replacement map contains the following entries:
3.32/1.60	
3.32/1.60	terms: {1}
3.32/1.60	cons: {1}
3.32/1.60	recip: {1}
3.32/1.60	sqr: {1}
3.32/1.60	s: {1}
3.32/1.60	0: empty set
3.32/1.60	add: {1, 2}
3.32/1.60	dbl: {1}
3.32/1.60	first: {1, 2}
3.32/1.60	nil: empty set
3.32/1.60	half: {1}
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(1) CSDependencyPairsProof (EQUIVALENT)
3.32/1.60	Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(2)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1, SQR_1, TERMS_1, ADD_2, DBL_1, HALF_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The ordinary context-sensitive dependency pairs DP_o are:
3.32/1.60	   TERMS(N) -> SQR(N)
3.32/1.60	   SQR(s(X)) -> ADD(sqr(X), dbl(X))
3.32/1.60	   SQR(s(X)) -> SQR(X)
3.32/1.60	   SQR(s(X)) -> DBL(X)
3.32/1.60	   DBL(s(X)) -> DBL(X)
3.32/1.60	   ADD(s(X), Y) -> ADD(X, Y)
3.32/1.60	   HALF(s(s(X))) -> HALF(X)
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(3) QCSDependencyGraphProof (EQUIVALENT)
3.32/1.60	The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 4 SCCs with 3 less nodes.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(4)
3.32/1.60	Complex Obligation (AND)
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(5)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1, HALF_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	
3.32/1.60	   HALF(s(s(X))) -> HALF(X)
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(6) QCSDPSubtermProof (EQUIVALENT)
3.32/1.60	We use the subterm processor [DA_EMMES].
3.32/1.60	
3.32/1.60	
3.32/1.60	The following pairs can be oriented strictly and are deleted.
3.32/1.60	
3.32/1.60	   HALF(s(s(X))) -> HALF(X)
3.32/1.60	The remaining pairs can at least be oriented weakly.
3.32/1.60	none
3.32/1.60	Used ordering:  Combined order from the following AFS and order.
3.32/1.60	HALF(x1)  =  x1
3.32/1.60	
3.32/1.60	
3.32/1.60	Subterm Order
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(7)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	none
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(8) PIsEmptyProof (EQUIVALENT)
3.32/1.60	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(9)
3.32/1.60	YES
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(10)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1, DBL_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	
3.32/1.60	   DBL(s(X)) -> DBL(X)
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(11) QCSDPSubtermProof (EQUIVALENT)
3.32/1.60	We use the subterm processor [DA_EMMES].
3.32/1.60	
3.32/1.60	
3.32/1.60	The following pairs can be oriented strictly and are deleted.
3.32/1.60	
3.32/1.60	   DBL(s(X)) -> DBL(X)
3.32/1.60	The remaining pairs can at least be oriented weakly.
3.32/1.60	none
3.32/1.60	Used ordering:  Combined order from the following AFS and order.
3.32/1.60	DBL(x1)  =  x1
3.32/1.60	
3.32/1.60	
3.32/1.60	Subterm Order
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(12)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	none
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(13) PIsEmptyProof (EQUIVALENT)
3.32/1.60	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(14)
3.32/1.60	YES
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(15)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1, ADD_2} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	
3.32/1.60	   ADD(s(X), Y) -> ADD(X, Y)
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(16) QCSDPSubtermProof (EQUIVALENT)
3.32/1.60	We use the subterm processor [DA_EMMES].
3.32/1.60	
3.32/1.60	
3.32/1.60	The following pairs can be oriented strictly and are deleted.
3.32/1.60	
3.32/1.60	   ADD(s(X), Y) -> ADD(X, Y)
3.32/1.60	The remaining pairs can at least be oriented weakly.
3.32/1.60	none
3.32/1.60	Used ordering:  Combined order from the following AFS and order.
3.32/1.60	ADD(x1, x2)  =  x1
3.32/1.60	
3.32/1.60	
3.32/1.60	Subterm Order
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(17)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	none
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(18) PIsEmptyProof (EQUIVALENT)
3.32/1.60	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(19)
3.32/1.60	YES
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(20)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1, SQR_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	
3.32/1.60	   SQR(s(X)) -> SQR(X)
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(21) QCSDPSubtermProof (EQUIVALENT)
3.32/1.60	We use the subterm processor [DA_EMMES].
3.32/1.60	
3.32/1.60	
3.32/1.60	The following pairs can be oriented strictly and are deleted.
3.32/1.60	
3.32/1.60	   SQR(s(X)) -> SQR(X)
3.32/1.60	The remaining pairs can at least be oriented weakly.
3.32/1.60	none
3.32/1.60	Used ordering:  Combined order from the following AFS and order.
3.32/1.60	SQR(x1)  =  x1
3.32/1.60	
3.32/1.60	
3.32/1.60	Subterm Order
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(22)
3.32/1.60	Obligation:
3.32/1.60	Q-restricted context-sensitive dependency pair problem:
3.32/1.60	The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, half_1} are replacing on all positions.
3.32/1.60	For all symbols f in {cons_2} we have mu(f) = {1}.
3.32/1.60	
3.32/1.60	The TRS P consists of the following rules:
3.32/1.60	none
3.32/1.60	
3.32/1.60	The TRS R consists of the following rules:
3.32/1.60	
3.32/1.60	   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
3.32/1.60	   sqr(0) -> 0
3.32/1.60	   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
3.32/1.60	   dbl(0) -> 0
3.32/1.60	   dbl(s(X)) -> s(s(dbl(X)))
3.32/1.60	   add(0, X) -> X
3.32/1.60	   add(s(X), Y) -> s(add(X, Y))
3.32/1.60	   first(0, X) -> nil
3.32/1.60	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.32/1.60	   half(0) -> 0
3.32/1.60	   half(s(0)) -> 0
3.32/1.60	   half(s(s(X))) -> s(half(X))
3.32/1.60	   half(dbl(X)) -> X
3.32/1.60	
3.32/1.60	Q is empty.
3.32/1.60	
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(23) PIsEmptyProof (EQUIVALENT)
3.32/1.60	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.32/1.60	----------------------------------------
3.32/1.60	
3.32/1.60	(24)
3.32/1.60	YES
3.32/1.64	EOF