3.67/1.83	YES
3.67/1.84	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
3.67/1.84	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.67/1.84	
3.67/1.84	
3.67/1.84	Termination w.r.t. Q of the given QTRS could be proven:
3.67/1.84	
3.67/1.84	(0) QTRS
3.67/1.84	(1) QTRSToCSRProof [SOUND, 0 ms]
3.67/1.84	(2) CSR
3.67/1.84	(3) CSRInnermostProof [EQUIVALENT, 0 ms]
3.67/1.84	(4) CSR
3.67/1.84	(5) CSDependencyPairsProof [EQUIVALENT, 11 ms]
3.67/1.84	(6) QCSDP
3.67/1.84	(7) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
3.67/1.84	(8) AND
3.67/1.84	    (9) QCSDP
3.67/1.84	        (10) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.67/1.84	        (11) QCSDP
3.67/1.84	        (12) PIsEmptyProof [EQUIVALENT, 0 ms]
3.67/1.84	        (13) YES
3.67/1.84	    (14) QCSDP
3.67/1.84	        (15) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.67/1.84	        (16) QCSDP
3.67/1.84	        (17) PIsEmptyProof [EQUIVALENT, 0 ms]
3.67/1.84	        (18) YES
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(0)
3.67/1.84	Obligation:
3.67/1.84	Q restricted rewrite system:
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   active(from(X)) -> mark(cons(X, from(s(X))))
3.67/1.84	   active(head(cons(X, XS))) -> mark(X)
3.67/1.84	   active(2nd(cons(X, XS))) -> mark(head(XS))
3.67/1.84	   active(take(0, XS)) -> mark(nil)
3.67/1.84	   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
3.67/1.84	   active(sel(0, cons(X, XS))) -> mark(X)
3.67/1.84	   active(sel(s(N), cons(X, XS))) -> mark(sel(N, XS))
3.67/1.84	   active(from(X)) -> from(active(X))
3.67/1.84	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.67/1.84	   active(s(X)) -> s(active(X))
3.67/1.84	   active(head(X)) -> head(active(X))
3.67/1.84	   active(2nd(X)) -> 2nd(active(X))
3.67/1.84	   active(take(X1, X2)) -> take(active(X1), X2)
3.67/1.84	   active(take(X1, X2)) -> take(X1, active(X2))
3.67/1.84	   active(sel(X1, X2)) -> sel(active(X1), X2)
3.67/1.84	   active(sel(X1, X2)) -> sel(X1, active(X2))
3.67/1.84	   from(mark(X)) -> mark(from(X))
3.67/1.84	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.67/1.84	   s(mark(X)) -> mark(s(X))
3.67/1.84	   head(mark(X)) -> mark(head(X))
3.67/1.84	   2nd(mark(X)) -> mark(2nd(X))
3.67/1.84	   take(mark(X1), X2) -> mark(take(X1, X2))
3.67/1.84	   take(X1, mark(X2)) -> mark(take(X1, X2))
3.67/1.84	   sel(mark(X1), X2) -> mark(sel(X1, X2))
3.67/1.84	   sel(X1, mark(X2)) -> mark(sel(X1, X2))
3.67/1.84	   proper(from(X)) -> from(proper(X))
3.67/1.84	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.67/1.84	   proper(s(X)) -> s(proper(X))
3.67/1.84	   proper(head(X)) -> head(proper(X))
3.67/1.84	   proper(2nd(X)) -> 2nd(proper(X))
3.67/1.84	   proper(take(X1, X2)) -> take(proper(X1), proper(X2))
3.67/1.84	   proper(0) -> ok(0)
3.67/1.84	   proper(nil) -> ok(nil)
3.67/1.84	   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
3.67/1.84	   from(ok(X)) -> ok(from(X))
3.67/1.84	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.67/1.84	   s(ok(X)) -> ok(s(X))
3.67/1.84	   head(ok(X)) -> ok(head(X))
3.67/1.84	   2nd(ok(X)) -> ok(2nd(X))
3.67/1.84	   take(ok(X1), ok(X2)) -> ok(take(X1, X2))
3.67/1.84	   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
3.67/1.84	   top(mark(X)) -> top(proper(X))
3.67/1.84	   top(ok(X)) -> top(active(X))
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   active(from(x0))
3.67/1.84	   active(cons(x0, x1))
3.67/1.84	   active(s(x0))
3.67/1.84	   active(head(x0))
3.67/1.84	   active(2nd(x0))
3.67/1.84	   active(take(x0, x1))
3.67/1.84	   active(sel(x0, x1))
3.67/1.84	   from(mark(x0))
3.67/1.84	   cons(mark(x0), x1)
3.67/1.84	   s(mark(x0))
3.67/1.84	   head(mark(x0))
3.67/1.84	   2nd(mark(x0))
3.67/1.84	   take(mark(x0), x1)
3.67/1.84	   take(x0, mark(x1))
3.67/1.84	   sel(mark(x0), x1)
3.67/1.84	   sel(x0, mark(x1))
3.67/1.84	   proper(from(x0))
3.67/1.84	   proper(cons(x0, x1))
3.67/1.84	   proper(s(x0))
3.67/1.84	   proper(head(x0))
3.67/1.84	   proper(2nd(x0))
3.67/1.84	   proper(take(x0, x1))
3.67/1.84	   proper(0)
3.67/1.84	   proper(nil)
3.67/1.84	   proper(sel(x0, x1))
3.67/1.84	   from(ok(x0))
3.67/1.84	   cons(ok(x0), ok(x1))
3.67/1.84	   s(ok(x0))
3.67/1.84	   head(ok(x0))
3.67/1.84	   2nd(ok(x0))
3.67/1.84	   take(ok(x0), ok(x1))
3.67/1.84	   sel(ok(x0), ok(x1))
3.67/1.84	   top(mark(x0))
3.67/1.84	   top(ok(x0))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(1) QTRSToCSRProof (SOUND)
3.67/1.84	The following Q TRS is given: Q restricted rewrite system:
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   active(from(X)) -> mark(cons(X, from(s(X))))
3.67/1.84	   active(head(cons(X, XS))) -> mark(X)
3.67/1.84	   active(2nd(cons(X, XS))) -> mark(head(XS))
3.67/1.84	   active(take(0, XS)) -> mark(nil)
3.67/1.84	   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
3.67/1.84	   active(sel(0, cons(X, XS))) -> mark(X)
3.67/1.84	   active(sel(s(N), cons(X, XS))) -> mark(sel(N, XS))
3.67/1.84	   active(from(X)) -> from(active(X))
3.67/1.84	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.67/1.84	   active(s(X)) -> s(active(X))
3.67/1.84	   active(head(X)) -> head(active(X))
3.67/1.84	   active(2nd(X)) -> 2nd(active(X))
3.67/1.84	   active(take(X1, X2)) -> take(active(X1), X2)
3.67/1.84	   active(take(X1, X2)) -> take(X1, active(X2))
3.67/1.84	   active(sel(X1, X2)) -> sel(active(X1), X2)
3.67/1.84	   active(sel(X1, X2)) -> sel(X1, active(X2))
3.67/1.84	   from(mark(X)) -> mark(from(X))
3.67/1.84	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.67/1.84	   s(mark(X)) -> mark(s(X))
3.67/1.84	   head(mark(X)) -> mark(head(X))
3.67/1.84	   2nd(mark(X)) -> mark(2nd(X))
3.67/1.84	   take(mark(X1), X2) -> mark(take(X1, X2))
3.67/1.84	   take(X1, mark(X2)) -> mark(take(X1, X2))
3.67/1.84	   sel(mark(X1), X2) -> mark(sel(X1, X2))
3.67/1.84	   sel(X1, mark(X2)) -> mark(sel(X1, X2))
3.67/1.84	   proper(from(X)) -> from(proper(X))
3.67/1.84	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.67/1.84	   proper(s(X)) -> s(proper(X))
3.67/1.84	   proper(head(X)) -> head(proper(X))
3.67/1.84	   proper(2nd(X)) -> 2nd(proper(X))
3.67/1.84	   proper(take(X1, X2)) -> take(proper(X1), proper(X2))
3.67/1.84	   proper(0) -> ok(0)
3.67/1.84	   proper(nil) -> ok(nil)
3.67/1.84	   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
3.67/1.84	   from(ok(X)) -> ok(from(X))
3.67/1.84	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.67/1.84	   s(ok(X)) -> ok(s(X))
3.67/1.84	   head(ok(X)) -> ok(head(X))
3.67/1.84	   2nd(ok(X)) -> ok(2nd(X))
3.67/1.84	   take(ok(X1), ok(X2)) -> ok(take(X1, X2))
3.67/1.84	   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
3.67/1.84	   top(mark(X)) -> top(proper(X))
3.67/1.84	   top(ok(X)) -> top(active(X))
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   active(from(x0))
3.67/1.84	   active(cons(x0, x1))
3.67/1.84	   active(s(x0))
3.67/1.84	   active(head(x0))
3.67/1.84	   active(2nd(x0))
3.67/1.84	   active(take(x0, x1))
3.67/1.84	   active(sel(x0, x1))
3.67/1.84	   from(mark(x0))
3.67/1.84	   cons(mark(x0), x1)
3.67/1.84	   s(mark(x0))
3.67/1.84	   head(mark(x0))
3.67/1.84	   2nd(mark(x0))
3.67/1.84	   take(mark(x0), x1)
3.67/1.84	   take(x0, mark(x1))
3.67/1.84	   sel(mark(x0), x1)
3.67/1.84	   sel(x0, mark(x1))
3.67/1.84	   proper(from(x0))
3.67/1.84	   proper(cons(x0, x1))
3.67/1.84	   proper(s(x0))
3.67/1.84	   proper(head(x0))
3.67/1.84	   proper(2nd(x0))
3.67/1.84	   proper(take(x0, x1))
3.67/1.84	   proper(0)
3.67/1.84	   proper(nil)
3.67/1.84	   proper(sel(x0, x1))
3.67/1.84	   from(ok(x0))
3.67/1.84	   cons(ok(x0), ok(x1))
3.67/1.84	   s(ok(x0))
3.67/1.84	   head(ok(x0))
3.67/1.84	   2nd(ok(x0))
3.67/1.84	   take(ok(x0), ok(x1))
3.67/1.84	   sel(ok(x0), ok(x1))
3.67/1.84	   top(mark(x0))
3.67/1.84	   top(ok(x0))
3.67/1.84	
3.67/1.84	Special symbols used for the transformation (see [GM04]):
3.67/1.84	top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
3.67/1.84	The replacement map contains the following entries:
3.67/1.84	
3.67/1.84	from: {1}
3.67/1.84	cons: {1}
3.67/1.84	s: {1}
3.67/1.84	head: {1}
3.67/1.84	2nd: {1}
3.67/1.84	take: {1, 2}
3.67/1.84	0: empty set
3.67/1.84	nil: empty set
3.67/1.84	sel: {1, 2}
3.67/1.84	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.
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(2)
3.67/1.84	Obligation:
3.67/1.84	Context-sensitive rewrite system:
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The replacement map contains the following entries:
3.67/1.84	
3.67/1.84	from: {1}
3.67/1.84	cons: {1}
3.67/1.84	s: {1}
3.67/1.84	head: {1}
3.67/1.84	2nd: {1}
3.67/1.84	take: {1, 2}
3.67/1.84	0: empty set
3.67/1.84	nil: empty set
3.67/1.84	sel: {1, 2}
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(3) CSRInnermostProof (EQUIVALENT)
3.67/1.84	The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(4)
3.67/1.84	Obligation:
3.67/1.84	Context-sensitive rewrite system:
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The replacement map contains the following entries:
3.67/1.84	
3.67/1.84	from: {1}
3.67/1.84	cons: {1}
3.67/1.84	s: {1}
3.67/1.84	head: {1}
3.67/1.84	2nd: {1}
3.67/1.84	take: {1, 2}
3.67/1.84	0: empty set
3.67/1.84	nil: empty set
3.67/1.84	sel: {1, 2}
3.67/1.84	
3.67/1.84	
3.67/1.84	Innermost Strategy.
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(5) CSDependencyPairsProof (EQUIVALENT)
3.67/1.84	Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(6)
3.67/1.84	Obligation:
3.67/1.84	Q-restricted context-sensitive dependency pair problem:
3.67/1.84	The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2, HEAD_1, 2ND_1, SEL_2, FROM_1, TAKE_2} are replacing on all positions.
3.67/1.84	For all symbols f in {cons_2} we have mu(f) = {1}.
3.67/1.84	The symbols in {U_1} are not replacing on any position.
3.67/1.84	
3.67/1.84	The ordinary context-sensitive dependency pairs DP_o are:
3.67/1.84	   2ND(cons(X, XS)) -> HEAD(XS)
3.67/1.84	   SEL(s(N), cons(X, XS)) -> SEL(N, XS)
3.67/1.84	
3.67/1.84	The collapsing dependency pairs are DP_c:
3.67/1.84	   2ND(cons(X, XS)) -> XS
3.67/1.84	   SEL(s(N), cons(X, XS)) -> XS
3.67/1.84	
3.67/1.84	
3.67/1.84	The hidden terms of R are:
3.67/1.84	
3.67/1.84	   from(s(x0))
3.67/1.84	   take(x0, x1)
3.67/1.84	
3.67/1.84	Every hiding context is built from:
3.67/1.84	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@62944c59
3.67/1.84	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@389324ca
3.67/1.84	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@46d90411
3.67/1.84	
3.67/1.84	Hence, the new unhiding pairs DP_u are :
3.67/1.84	   2ND(cons(X, XS)) -> U(XS)
3.67/1.84	   SEL(s(N), cons(X, XS)) -> U(XS)
3.67/1.84	   U(s(x_0)) -> U(x_0)
3.67/1.84	   U(from(x_0)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_1)
3.67/1.84	   U(from(s(x0))) -> FROM(s(x0))
3.67/1.84	   U(take(x0, x1)) -> TAKE(x0, x1)
3.67/1.84	
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   from(x0)
3.67/1.84	   head(cons(x0, x1))
3.67/1.84	   2nd(cons(x0, x1))
3.67/1.84	   take(0, x0)
3.67/1.84	   take(s(x0), cons(x1, x2))
3.67/1.84	   sel(0, cons(x0, x1))
3.67/1.84	   sel(s(x0), cons(x1, x2))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(7) QCSDependencyGraphProof (EQUIVALENT)
3.67/1.84	The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 2 SCCs with 4 less nodes.
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(8)
3.67/1.84	Complex Obligation (AND)
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(9)
3.67/1.84	Obligation:
3.67/1.84	Q-restricted context-sensitive dependency pair problem:
3.67/1.84	The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
3.67/1.84	For all symbols f in {cons_2} we have mu(f) = {1}.
3.67/1.84	The symbols in {U_1} are not replacing on any position.
3.67/1.84	
3.67/1.84	The TRS P consists of the following rules:
3.67/1.84	
3.67/1.84	   U(s(x_0)) -> U(x_0)
3.67/1.84	   U(from(x_0)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_1)
3.67/1.84	
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   from(x0)
3.67/1.84	   head(cons(x0, x1))
3.67/1.84	   2nd(cons(x0, x1))
3.67/1.84	   take(0, x0)
3.67/1.84	   take(s(x0), cons(x1, x2))
3.67/1.84	   sel(0, cons(x0, x1))
3.67/1.84	   sel(s(x0), cons(x1, x2))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(10) QCSDPSubtermProof (EQUIVALENT)
3.67/1.84	We use the subterm processor [DA_EMMES].
3.67/1.84	
3.67/1.84	
3.67/1.84	The following pairs can be oriented strictly and are deleted.
3.67/1.84	
3.67/1.84	   U(s(x_0)) -> U(x_0)
3.67/1.84	   U(from(x_0)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_0)
3.67/1.84	   U(take(x_0, x_1)) -> U(x_1)
3.67/1.84	The remaining pairs can at least be oriented weakly.
3.67/1.84	none
3.67/1.84	Used ordering:  Combined order from the following AFS and order.
3.67/1.84	U(x1)  =  x1
3.67/1.84	
3.67/1.84	
3.67/1.84	Subterm Order
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(11)
3.67/1.84	Obligation:
3.67/1.84	Q-restricted context-sensitive dependency pair problem:
3.67/1.84	The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
3.67/1.84	For all symbols f in {cons_2} we have mu(f) = {1}.
3.67/1.84	
3.67/1.84	The TRS P consists of the following rules:
3.67/1.84	none
3.67/1.84	
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   from(x0)
3.67/1.84	   head(cons(x0, x1))
3.67/1.84	   2nd(cons(x0, x1))
3.67/1.84	   take(0, x0)
3.67/1.84	   take(s(x0), cons(x1, x2))
3.67/1.84	   sel(0, cons(x0, x1))
3.67/1.84	   sel(s(x0), cons(x1, x2))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(12) PIsEmptyProof (EQUIVALENT)
3.67/1.84	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(13)
3.67/1.84	YES
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(14)
3.67/1.84	Obligation:
3.67/1.84	Q-restricted context-sensitive dependency pair problem:
3.67/1.84	The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2, SEL_2} are replacing on all positions.
3.67/1.84	For all symbols f in {cons_2} we have mu(f) = {1}.
3.67/1.84	
3.67/1.84	The TRS P consists of the following rules:
3.67/1.84	
3.67/1.84	   SEL(s(N), cons(X, XS)) -> SEL(N, XS)
3.67/1.84	
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   from(x0)
3.67/1.84	   head(cons(x0, x1))
3.67/1.84	   2nd(cons(x0, x1))
3.67/1.84	   take(0, x0)
3.67/1.84	   take(s(x0), cons(x1, x2))
3.67/1.84	   sel(0, cons(x0, x1))
3.67/1.84	   sel(s(x0), cons(x1, x2))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(15) QCSDPSubtermProof (EQUIVALENT)
3.67/1.84	We use the subterm processor [DA_EMMES].
3.67/1.84	
3.67/1.84	
3.67/1.84	The following pairs can be oriented strictly and are deleted.
3.67/1.84	
3.67/1.84	   SEL(s(N), cons(X, XS)) -> SEL(N, XS)
3.67/1.84	The remaining pairs can at least be oriented weakly.
3.67/1.84	none
3.67/1.84	Used ordering:  Combined order from the following AFS and order.
3.67/1.84	SEL(x1, x2)  =  x1
3.67/1.84	
3.67/1.84	
3.67/1.84	Subterm Order
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(16)
3.67/1.84	Obligation:
3.67/1.84	Q-restricted context-sensitive dependency pair problem:
3.67/1.84	The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
3.67/1.84	For all symbols f in {cons_2} we have mu(f) = {1}.
3.67/1.84	
3.67/1.84	The TRS P consists of the following rules:
3.67/1.84	none
3.67/1.84	
3.67/1.84	The TRS R consists of the following rules:
3.67/1.84	
3.67/1.84	   from(X) -> cons(X, from(s(X)))
3.67/1.84	   head(cons(X, XS)) -> X
3.67/1.84	   2nd(cons(X, XS)) -> head(XS)
3.67/1.84	   take(0, XS) -> nil
3.67/1.84	   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
3.67/1.84	   sel(0, cons(X, XS)) -> X
3.67/1.84	   sel(s(N), cons(X, XS)) -> sel(N, XS)
3.67/1.84	
3.67/1.84	The set Q consists of the following terms:
3.67/1.84	
3.67/1.84	   from(x0)
3.67/1.84	   head(cons(x0, x1))
3.67/1.84	   2nd(cons(x0, x1))
3.67/1.84	   take(0, x0)
3.67/1.84	   take(s(x0), cons(x1, x2))
3.67/1.84	   sel(0, cons(x0, x1))
3.67/1.84	   sel(s(x0), cons(x1, x2))
3.67/1.84	
3.67/1.84	
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(17) PIsEmptyProof (EQUIVALENT)
3.67/1.84	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.67/1.84	----------------------------------------
3.67/1.84	
3.67/1.84	(18)
3.67/1.84	YES
3.91/1.88	EOF