3.46/1.78	YES
3.46/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.46/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.46/1.79	
3.46/1.79	
3.46/1.79	Termination w.r.t. Q of the given QTRS could be proven:
3.46/1.79	
3.46/1.79	(0) QTRS
3.46/1.79	(1) QTRSToCSRProof [SOUND, 0 ms]
3.46/1.79	(2) CSR
3.46/1.79	(3) CSRInnermostProof [EQUIVALENT, 0 ms]
3.46/1.79	(4) CSR
3.46/1.79	(5) CSDependencyPairsProof [EQUIVALENT, 0 ms]
3.46/1.79	(6) QCSDP
3.46/1.79	(7) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
3.46/1.79	(8) AND
3.46/1.79	    (9) QCSDP
3.46/1.79	        (10) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.46/1.79	        (11) QCSDP
3.46/1.79	        (12) PIsEmptyProof [EQUIVALENT, 0 ms]
3.46/1.79	        (13) YES
3.46/1.79	    (14) QCSDP
3.46/1.79	        (15) QCSDPSubtermProof [EQUIVALENT, 0 ms]
3.46/1.79	        (16) QCSDP
3.46/1.79	        (17) PIsEmptyProof [EQUIVALENT, 0 ms]
3.46/1.79	        (18) YES
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(0)
3.46/1.79	Obligation:
3.46/1.79	Q restricted rewrite system:
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   active(from(X)) -> mark(cons(X, from(s(X))))
3.46/1.79	   active(first(0, Z)) -> mark(nil)
3.46/1.79	   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
3.46/1.79	   active(sel(0, cons(X, Z))) -> mark(X)
3.46/1.79	   active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
3.46/1.79	   active(from(X)) -> from(active(X))
3.46/1.79	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.46/1.79	   active(s(X)) -> s(active(X))
3.46/1.79	   active(first(X1, X2)) -> first(active(X1), X2)
3.46/1.79	   active(first(X1, X2)) -> first(X1, active(X2))
3.46/1.79	   active(sel(X1, X2)) -> sel(active(X1), X2)
3.46/1.79	   active(sel(X1, X2)) -> sel(X1, active(X2))
3.46/1.79	   from(mark(X)) -> mark(from(X))
3.46/1.79	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.46/1.79	   s(mark(X)) -> mark(s(X))
3.46/1.79	   first(mark(X1), X2) -> mark(first(X1, X2))
3.46/1.79	   first(X1, mark(X2)) -> mark(first(X1, X2))
3.46/1.79	   sel(mark(X1), X2) -> mark(sel(X1, X2))
3.46/1.79	   sel(X1, mark(X2)) -> mark(sel(X1, X2))
3.46/1.79	   proper(from(X)) -> from(proper(X))
3.46/1.79	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.46/1.79	   proper(s(X)) -> s(proper(X))
3.46/1.79	   proper(first(X1, X2)) -> first(proper(X1), proper(X2))
3.46/1.79	   proper(0) -> ok(0)
3.46/1.79	   proper(nil) -> ok(nil)
3.46/1.79	   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
3.46/1.79	   from(ok(X)) -> ok(from(X))
3.46/1.79	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.46/1.79	   s(ok(X)) -> ok(s(X))
3.46/1.79	   first(ok(X1), ok(X2)) -> ok(first(X1, X2))
3.46/1.79	   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
3.46/1.79	   top(mark(X)) -> top(proper(X))
3.46/1.79	   top(ok(X)) -> top(active(X))
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   active(from(x0))
3.46/1.79	   active(cons(x0, x1))
3.46/1.79	   active(s(x0))
3.46/1.79	   active(first(x0, x1))
3.46/1.79	   active(sel(x0, x1))
3.46/1.79	   from(mark(x0))
3.46/1.79	   cons(mark(x0), x1)
3.46/1.79	   s(mark(x0))
3.46/1.79	   first(mark(x0), x1)
3.46/1.79	   first(x0, mark(x1))
3.46/1.79	   sel(mark(x0), x1)
3.46/1.79	   sel(x0, mark(x1))
3.46/1.79	   proper(from(x0))
3.46/1.79	   proper(cons(x0, x1))
3.46/1.79	   proper(s(x0))
3.46/1.79	   proper(first(x0, x1))
3.46/1.79	   proper(0)
3.46/1.79	   proper(nil)
3.46/1.79	   proper(sel(x0, x1))
3.46/1.79	   from(ok(x0))
3.46/1.79	   cons(ok(x0), ok(x1))
3.46/1.79	   s(ok(x0))
3.46/1.79	   first(ok(x0), ok(x1))
3.46/1.79	   sel(ok(x0), ok(x1))
3.46/1.79	   top(mark(x0))
3.46/1.79	   top(ok(x0))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(1) QTRSToCSRProof (SOUND)
3.46/1.79	The following Q TRS is given: Q restricted rewrite system:
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   active(from(X)) -> mark(cons(X, from(s(X))))
3.46/1.79	   active(first(0, Z)) -> mark(nil)
3.46/1.79	   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
3.46/1.79	   active(sel(0, cons(X, Z))) -> mark(X)
3.46/1.79	   active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
3.46/1.79	   active(from(X)) -> from(active(X))
3.46/1.79	   active(cons(X1, X2)) -> cons(active(X1), X2)
3.46/1.79	   active(s(X)) -> s(active(X))
3.46/1.79	   active(first(X1, X2)) -> first(active(X1), X2)
3.46/1.79	   active(first(X1, X2)) -> first(X1, active(X2))
3.46/1.79	   active(sel(X1, X2)) -> sel(active(X1), X2)
3.46/1.79	   active(sel(X1, X2)) -> sel(X1, active(X2))
3.46/1.79	   from(mark(X)) -> mark(from(X))
3.46/1.79	   cons(mark(X1), X2) -> mark(cons(X1, X2))
3.46/1.79	   s(mark(X)) -> mark(s(X))
3.46/1.79	   first(mark(X1), X2) -> mark(first(X1, X2))
3.46/1.79	   first(X1, mark(X2)) -> mark(first(X1, X2))
3.46/1.79	   sel(mark(X1), X2) -> mark(sel(X1, X2))
3.46/1.79	   sel(X1, mark(X2)) -> mark(sel(X1, X2))
3.46/1.79	   proper(from(X)) -> from(proper(X))
3.46/1.79	   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
3.46/1.79	   proper(s(X)) -> s(proper(X))
3.46/1.79	   proper(first(X1, X2)) -> first(proper(X1), proper(X2))
3.46/1.79	   proper(0) -> ok(0)
3.46/1.79	   proper(nil) -> ok(nil)
3.46/1.79	   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
3.46/1.79	   from(ok(X)) -> ok(from(X))
3.46/1.79	   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
3.46/1.79	   s(ok(X)) -> ok(s(X))
3.46/1.79	   first(ok(X1), ok(X2)) -> ok(first(X1, X2))
3.46/1.79	   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
3.46/1.79	   top(mark(X)) -> top(proper(X))
3.46/1.79	   top(ok(X)) -> top(active(X))
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   active(from(x0))
3.46/1.79	   active(cons(x0, x1))
3.46/1.79	   active(s(x0))
3.46/1.79	   active(first(x0, x1))
3.46/1.79	   active(sel(x0, x1))
3.46/1.79	   from(mark(x0))
3.46/1.79	   cons(mark(x0), x1)
3.46/1.79	   s(mark(x0))
3.46/1.79	   first(mark(x0), x1)
3.46/1.79	   first(x0, mark(x1))
3.46/1.79	   sel(mark(x0), x1)
3.46/1.79	   sel(x0, mark(x1))
3.46/1.79	   proper(from(x0))
3.46/1.79	   proper(cons(x0, x1))
3.46/1.79	   proper(s(x0))
3.46/1.79	   proper(first(x0, x1))
3.46/1.79	   proper(0)
3.46/1.79	   proper(nil)
3.46/1.79	   proper(sel(x0, x1))
3.46/1.79	   from(ok(x0))
3.46/1.79	   cons(ok(x0), ok(x1))
3.46/1.79	   s(ok(x0))
3.46/1.79	   first(ok(x0), ok(x1))
3.46/1.79	   sel(ok(x0), ok(x1))
3.46/1.79	   top(mark(x0))
3.46/1.79	   top(ok(x0))
3.46/1.79	
3.46/1.79	Special symbols used for the transformation (see [GM04]):
3.46/1.79	top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
3.46/1.79	The replacement map contains the following entries:
3.46/1.79	
3.46/1.79	from: {1}
3.46/1.79	cons: {1}
3.46/1.79	s: {1}
3.46/1.79	first: {1, 2}
3.46/1.79	0: empty set
3.46/1.79	nil: empty set
3.46/1.79	sel: {1, 2}
3.46/1.79	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.
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(2)
3.46/1.79	Obligation:
3.46/1.79	Context-sensitive rewrite system:
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The replacement map contains the following entries:
3.46/1.79	
3.46/1.79	from: {1}
3.46/1.79	cons: {1}
3.46/1.79	s: {1}
3.46/1.79	first: {1, 2}
3.46/1.79	0: empty set
3.46/1.79	nil: empty set
3.46/1.79	sel: {1, 2}
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(3) CSRInnermostProof (EQUIVALENT)
3.46/1.79	The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(4)
3.46/1.79	Obligation:
3.46/1.79	Context-sensitive rewrite system:
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The replacement map contains the following entries:
3.46/1.79	
3.46/1.79	from: {1}
3.46/1.79	cons: {1}
3.46/1.79	s: {1}
3.46/1.79	first: {1, 2}
3.46/1.79	0: empty set
3.46/1.79	nil: empty set
3.46/1.79	sel: {1, 2}
3.46/1.79	
3.46/1.79	
3.46/1.79	Innermost Strategy.
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(5) CSDependencyPairsProof (EQUIVALENT)
3.46/1.79	Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(6)
3.46/1.79	Obligation:
3.46/1.79	Q-restricted context-sensitive dependency pair problem:
3.46/1.79	The symbols in {from_1, s_1, first_2, sel_2, SEL_2, FROM_1, FIRST_2} are replacing on all positions.
3.46/1.79	For all symbols f in {cons_2} we have mu(f) = {1}.
3.46/1.79	The symbols in {U_1} are not replacing on any position.
3.46/1.79	
3.46/1.79	The ordinary context-sensitive dependency pairs DP_o are:
3.46/1.79	   SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
3.46/1.79	
3.46/1.79	The collapsing dependency pairs are DP_c:
3.46/1.79	   SEL(s(X), cons(Y, Z)) -> Z
3.46/1.79	
3.46/1.79	
3.46/1.79	The hidden terms of R are:
3.46/1.79	
3.46/1.79	   from(s(x0))
3.46/1.79	   first(x0, x1)
3.46/1.79	
3.46/1.79	Every hiding context is built from:
3.46/1.79	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@1e3cf39e
3.46/1.79	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@d6ed18
3.46/1.79	   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@66fad4ed
3.46/1.79	
3.46/1.79	Hence, the new unhiding pairs DP_u are :
3.46/1.79	   SEL(s(X), cons(Y, Z)) -> U(Z)
3.46/1.79	   U(s(x_0)) -> U(x_0)
3.46/1.79	   U(from(x_0)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_1)
3.46/1.79	   U(from(s(x0))) -> FROM(s(x0))
3.46/1.79	   U(first(x0, x1)) -> FIRST(x0, x1)
3.46/1.79	
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   from(x0)
3.46/1.79	   first(0, x0)
3.46/1.79	   first(s(x0), cons(x1, x2))
3.46/1.79	   sel(0, cons(x0, x1))
3.46/1.79	   sel(s(x0), cons(x1, x2))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(7) QCSDependencyGraphProof (EQUIVALENT)
3.46/1.79	The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 2 SCCs with 3 less nodes.
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(8)
3.46/1.79	Complex Obligation (AND)
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(9)
3.46/1.79	Obligation:
3.46/1.79	Q-restricted context-sensitive dependency pair problem:
3.46/1.79	The symbols in {from_1, s_1, first_2, sel_2} are replacing on all positions.
3.46/1.79	For all symbols f in {cons_2} we have mu(f) = {1}.
3.46/1.79	The symbols in {U_1} are not replacing on any position.
3.46/1.79	
3.46/1.79	The TRS P consists of the following rules:
3.46/1.79	
3.46/1.79	   U(s(x_0)) -> U(x_0)
3.46/1.79	   U(from(x_0)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_1)
3.46/1.79	
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   from(x0)
3.46/1.79	   first(0, x0)
3.46/1.79	   first(s(x0), cons(x1, x2))
3.46/1.79	   sel(0, cons(x0, x1))
3.46/1.79	   sel(s(x0), cons(x1, x2))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(10) QCSDPSubtermProof (EQUIVALENT)
3.46/1.79	We use the subterm processor [DA_EMMES].
3.46/1.79	
3.46/1.79	
3.46/1.79	The following pairs can be oriented strictly and are deleted.
3.46/1.79	
3.46/1.79	   U(s(x_0)) -> U(x_0)
3.46/1.79	   U(from(x_0)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_0)
3.46/1.79	   U(first(x_0, x_1)) -> U(x_1)
3.46/1.79	The remaining pairs can at least be oriented weakly.
3.46/1.79	none
3.46/1.79	Used ordering:  Combined order from the following AFS and order.
3.46/1.79	U(x1)  =  x1
3.46/1.79	
3.46/1.79	
3.46/1.79	Subterm Order
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(11)
3.46/1.79	Obligation:
3.46/1.79	Q-restricted context-sensitive dependency pair problem:
3.46/1.79	The symbols in {from_1, s_1, first_2, sel_2} are replacing on all positions.
3.46/1.79	For all symbols f in {cons_2} we have mu(f) = {1}.
3.46/1.79	
3.46/1.79	The TRS P consists of the following rules:
3.46/1.79	none
3.46/1.79	
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   from(x0)
3.46/1.79	   first(0, x0)
3.46/1.79	   first(s(x0), cons(x1, x2))
3.46/1.79	   sel(0, cons(x0, x1))
3.46/1.79	   sel(s(x0), cons(x1, x2))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(12) PIsEmptyProof (EQUIVALENT)
3.46/1.79	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(13)
3.46/1.79	YES
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(14)
3.46/1.79	Obligation:
3.46/1.79	Q-restricted context-sensitive dependency pair problem:
3.46/1.79	The symbols in {from_1, s_1, first_2, sel_2, SEL_2} are replacing on all positions.
3.46/1.79	For all symbols f in {cons_2} we have mu(f) = {1}.
3.46/1.79	
3.46/1.79	The TRS P consists of the following rules:
3.46/1.79	
3.46/1.79	   SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
3.46/1.79	
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   from(x0)
3.46/1.79	   first(0, x0)
3.46/1.79	   first(s(x0), cons(x1, x2))
3.46/1.79	   sel(0, cons(x0, x1))
3.46/1.79	   sel(s(x0), cons(x1, x2))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(15) QCSDPSubtermProof (EQUIVALENT)
3.46/1.79	We use the subterm processor [DA_EMMES].
3.46/1.79	
3.46/1.79	
3.46/1.79	The following pairs can be oriented strictly and are deleted.
3.46/1.79	
3.46/1.79	   SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
3.46/1.79	The remaining pairs can at least be oriented weakly.
3.46/1.79	none
3.46/1.79	Used ordering:  Combined order from the following AFS and order.
3.46/1.79	SEL(x1, x2)  =  x1
3.46/1.79	
3.46/1.79	
3.46/1.79	Subterm Order
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(16)
3.46/1.79	Obligation:
3.46/1.79	Q-restricted context-sensitive dependency pair problem:
3.46/1.79	The symbols in {from_1, s_1, first_2, sel_2} are replacing on all positions.
3.46/1.79	For all symbols f in {cons_2} we have mu(f) = {1}.
3.46/1.79	
3.46/1.79	The TRS P consists of the following rules:
3.46/1.79	none
3.46/1.79	
3.46/1.79	The TRS R consists of the following rules:
3.46/1.79	
3.46/1.79	   from(X) -> cons(X, from(s(X)))
3.46/1.79	   first(0, Z) -> nil
3.46/1.79	   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
3.46/1.79	   sel(0, cons(X, Z)) -> X
3.46/1.79	   sel(s(X), cons(Y, Z)) -> sel(X, Z)
3.46/1.79	
3.46/1.79	The set Q consists of the following terms:
3.46/1.79	
3.46/1.79	   from(x0)
3.46/1.79	   first(0, x0)
3.46/1.79	   first(s(x0), cons(x1, x2))
3.46/1.79	   sel(0, cons(x0, x1))
3.46/1.79	   sel(s(x0), cons(x1, x2))
3.46/1.79	
3.46/1.79	
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(17) PIsEmptyProof (EQUIVALENT)
3.46/1.79	The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
3.46/1.79	----------------------------------------
3.46/1.79	
3.46/1.79	(18)
3.46/1.79	YES
3.71/1.82	EOF