3.68/1.51	YES
3.68/1.52	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.68/1.52	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.68/1.52	
3.68/1.52	
3.68/1.52	Termination of the given CSR could be proven:
3.68/1.52	
3.68/1.52	(0) CSR
3.68/1.52	(1) CSRInnermostProof [EQUIVALENT, 0 ms]
3.68/1.52	(2) CSR
3.68/1.52	(3) CSDependencyPairsProof [EQUIVALENT, 0 ms]
3.68/1.52	(4) QCSDP
3.68/1.52	(5) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
3.68/1.52	(6) QCSDP
3.68/1.52	(7) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.68/1.52	(8) QCSDP
3.68/1.52	(9) PIsEmptyProof [EQUIVALENT, 0 ms]
3.68/1.52	(10) YES
3.68/1.52	
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(0)
3.68/1.52	Obligation:
3.68/1.52	Context-sensitive rewrite system:
3.68/1.52	The TRS R consists of the following rules:
3.68/1.52	
3.68/1.52	   primes -> sieve(from(s(s(0))))
3.68/1.52	   from(X) -> cons(X, from(s(X)))
3.68/1.52	   head(cons(X, Y)) -> X
3.68/1.52	   tail(cons(X, Y)) -> Y
3.68/1.52	   if(true, X, Y) -> X
3.68/1.52	   if(false, X, Y) -> Y
3.68/1.52	   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
3.68/1.52	
3.68/1.52	The replacement map contains the following entries:
3.68/1.52	
3.68/1.52	primes: empty set
3.68/1.52	sieve: {1}
3.68/1.52	from: {1}
3.68/1.52	s: {1}
3.68/1.52	0: empty set
3.68/1.52	cons: {1}
3.68/1.52	head: {1}
3.68/1.52	tail: {1}
3.68/1.52	if: {1}
3.68/1.52	true: empty set
3.68/1.52	false: empty set
3.68/1.52	filter: {1, 2}
3.68/1.52	divides: {1, 2}
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(1) CSRInnermostProof (EQUIVALENT)
3.68/1.52	The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(2)
3.68/1.52	Obligation:
3.68/1.52	Context-sensitive rewrite system:
3.68/1.52	The TRS R consists of the following rules:
3.68/1.52	
3.68/1.52	   primes -> sieve(from(s(s(0))))
3.68/1.52	   from(X) -> cons(X, from(s(X)))
3.68/1.52	   head(cons(X, Y)) -> X
3.68/1.52	   tail(cons(X, Y)) -> Y
3.68/1.52	   if(true, X, Y) -> X
3.68/1.52	   if(false, X, Y) -> Y
3.68/1.52	   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
3.68/1.52	
3.68/1.52	The replacement map contains the following entries:
3.68/1.52	
3.68/1.52	primes: empty set
3.68/1.52	sieve: {1}
3.68/1.52	from: {1}
3.68/1.52	s: {1}
3.68/1.52	0: empty set
3.68/1.52	cons: {1}
3.68/1.52	head: {1}
3.68/1.52	tail: {1}
3.68/1.52	if: {1}
3.68/1.52	true: empty set
3.68/1.52	false: empty set
3.68/1.52	filter: {1, 2}
3.68/1.52	divides: {1, 2}
3.68/1.52	
3.68/1.52	
3.68/1.52	Innermost Strategy.
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(3) CSDependencyPairsProof (EQUIVALENT)
3.68/1.52	Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(4)
3.68/1.52	Obligation:
3.68/1.52	Q-restricted context-sensitive dependency pair problem:
3.68/1.52	The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2, SIEVE_1, FROM_1, FILTER_2, TAIL_1} are replacing on all positions.
3.68/1.52	For all symbols f in {cons_2, if_3, IF_3} we have mu(f) = {1}.
3.68/1.52	The symbols in {U_1} are not replacing on any position.
3.68/1.52	
3.68/1.52	The ordinary context-sensitive dependency pairs DP_o are:
3.68/1.52	   PRIMES -> SIEVE(from(s(s(0))))
3.68/1.52	   PRIMES -> FROM(s(s(0)))
3.68/1.52	   FILTER(s(s(X)), cons(Y, Z)) -> IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	
3.68/1.52	The collapsing dependency pairs are DP_c:
3.68/1.52	   TAIL(cons(X, Y)) -> Y
3.68/1.52	   IF(true, X, Y) -> X
3.68/1.52	   IF(false, X, Y) -> Y
3.68/1.52	
3.68/1.52	
3.68/1.52	The hidden terms of R are:
3.68/1.52	
3.68/1.52	   from(s(x0))
3.68/1.52	   filter(s(s(x0)), x1)
3.68/1.52	   filter(x0, sieve(x1))
3.68/1.52	   sieve(x0)
3.68/1.52	
3.68/1.52	Every hiding context is built from:
3.68/1.52	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@57a9d99c
3.68/1.52	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@35a48193
3.68/1.52	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@10c61f26
3.68/1.52	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@760b142a
3.68/1.52	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@6ff7564e
3.68/1.52	
3.68/1.52	Hence, the new unhiding pairs DP_u are :
3.68/1.52	   TAIL(cons(X, Y)) -> U(Y)
3.68/1.52	   IF(true, X, Y) -> U(X)
3.68/1.52	   IF(false, X, Y) -> U(Y)
3.68/1.52	   U(s(x_0)) -> U(x_0)
3.68/1.52	   U(from(x_0)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_1)
3.68/1.52	   U(sieve(x_0)) -> U(x_0)
3.68/1.52	   U(cons(x_0, x_1)) -> U(x_0)
3.68/1.52	   U(from(s(x0))) -> FROM(s(x0))
3.68/1.52	   U(filter(s(s(x0)), x1)) -> FILTER(s(s(x0)), x1)
3.68/1.52	   U(filter(x0, sieve(x1))) -> FILTER(x0, sieve(x1))
3.68/1.52	   U(sieve(x0)) -> SIEVE(x0)
3.68/1.52	
3.68/1.52	The TRS R consists of the following rules:
3.68/1.52	
3.68/1.52	   primes -> sieve(from(s(s(0))))
3.68/1.52	   from(X) -> cons(X, from(s(X)))
3.68/1.52	   head(cons(X, Y)) -> X
3.68/1.52	   tail(cons(X, Y)) -> Y
3.68/1.52	   if(true, X, Y) -> X
3.68/1.52	   if(false, X, Y) -> Y
3.68/1.52	   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
3.68/1.52	
3.68/1.52	The set Q consists of the following terms:
3.68/1.52	
3.68/1.52	   primes
3.68/1.52	   from(x0)
3.68/1.52	   head(cons(x0, x1))
3.68/1.52	   tail(cons(x0, x1))
3.68/1.52	   if(true, x0, x1)
3.68/1.52	   if(false, x0, x1)
3.68/1.52	   filter(s(s(x0)), cons(x1, x2))
3.68/1.52	   sieve(cons(x0, x1))
3.68/1.52	
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(5) QCSDependencyGraphProof (EQUIVALENT)
3.68/1.52	The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 1 SCC with 8 less nodes.
3.68/1.52	The rules FILTER(s(s(z0)), cons(z1, z2)) -> IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) and IF(true, x0, x1) -> U(x0) form no chain, because ECap^mu(IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1))))) = IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) does not unify with IF(true, x0, x1). The rules FILTER(s(s(z0)), cons(z1, z2)) -> IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) and IF(false, x0, x1) -> U(x1) form no chain, because ECap^mu(IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1))))) = IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) does not unify with IF(false, x0, x1). 
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(6)
3.68/1.52	Obligation:
3.68/1.52	Q-restricted context-sensitive dependency pair problem:
3.68/1.52	The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2} are replacing on all positions.
3.68/1.52	For all symbols f in {cons_2, if_3} we have mu(f) = {1}.
3.68/1.52	The symbols in {U_1} are not replacing on any position.
3.68/1.52	
3.68/1.52	The TRS P consists of the following rules:
3.68/1.52	
3.68/1.52	   U(s(x_0)) -> U(x_0)
3.68/1.52	   U(from(x_0)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_1)
3.68/1.52	   U(sieve(x_0)) -> U(x_0)
3.68/1.52	   U(cons(x_0, x_1)) -> U(x_0)
3.68/1.52	
3.68/1.52	The TRS R consists of the following rules:
3.68/1.52	
3.68/1.52	   primes -> sieve(from(s(s(0))))
3.68/1.52	   from(X) -> cons(X, from(s(X)))
3.68/1.52	   head(cons(X, Y)) -> X
3.68/1.52	   tail(cons(X, Y)) -> Y
3.68/1.52	   if(true, X, Y) -> X
3.68/1.52	   if(false, X, Y) -> Y
3.68/1.52	   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
3.68/1.52	
3.68/1.52	The set Q consists of the following terms:
3.68/1.52	
3.68/1.52	   primes
3.68/1.52	   from(x0)
3.68/1.52	   head(cons(x0, x1))
3.68/1.52	   tail(cons(x0, x1))
3.68/1.52	   if(true, x0, x1)
3.68/1.52	   if(false, x0, x1)
3.68/1.52	   filter(s(s(x0)), cons(x1, x2))
3.68/1.52	   sieve(cons(x0, x1))
3.68/1.52	
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(7) QCSDPSubtermProof (EQUIVALENT)
3.68/1.52	We use the subterm processor [DA_EMMES].
3.68/1.52	
3.68/1.52	
3.68/1.52	The following pairs can be oriented strictly and are deleted.
3.68/1.52	
3.68/1.52	   U(s(x_0)) -> U(x_0)
3.68/1.52	   U(from(x_0)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_0)
3.68/1.52	   U(filter(x_0, x_1)) -> U(x_1)
3.68/1.52	   U(sieve(x_0)) -> U(x_0)
3.68/1.52	   U(cons(x_0, x_1)) -> U(x_0)
3.68/1.52	The remaining pairs can at least be oriented weakly.
3.68/1.52	none
3.68/1.52	Used ordering:  Combined order from the following AFS and order.
3.68/1.52	U(x1)  =  x1
3.68/1.52	
3.68/1.52	
3.68/1.52	Subterm Order
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(8)
3.68/1.52	Obligation:
3.68/1.52	Q-restricted context-sensitive dependency pair problem:
3.68/1.52	The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2} are replacing on all positions.
3.68/1.52	For all symbols f in {cons_2, if_3} we have mu(f) = {1}.
3.68/1.52	
3.68/1.52	The TRS P consists of the following rules:
3.68/1.52	none
3.68/1.52	
3.68/1.52	The TRS R consists of the following rules:
3.68/1.52	
3.68/1.52	   primes -> sieve(from(s(s(0))))
3.68/1.52	   from(X) -> cons(X, from(s(X)))
3.68/1.52	   head(cons(X, Y)) -> X
3.68/1.52	   tail(cons(X, Y)) -> Y
3.68/1.52	   if(true, X, Y) -> X
3.68/1.52	   if(false, X, Y) -> Y
3.68/1.52	   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
3.68/1.52	   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
3.68/1.52	
3.68/1.52	The set Q consists of the following terms:
3.68/1.52	
3.68/1.52	   primes
3.68/1.52	   from(x0)
3.68/1.52	   head(cons(x0, x1))
3.68/1.52	   tail(cons(x0, x1))
3.68/1.52	   if(true, x0, x1)
3.68/1.52	   if(false, x0, x1)
3.68/1.52	   filter(s(s(x0)), cons(x1, x2))
3.68/1.52	   sieve(cons(x0, x1))
3.68/1.52	
3.68/1.52	
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(9) PIsEmptyProof (EQUIVALENT)
3.68/1.52	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.68/1.52	----------------------------------------
3.68/1.52	
3.68/1.52	(10)
3.68/1.52	YES
3.68/1.54	EOF