Logic_Programming 2019-04-02 08.22 pair #434324747

details

property	value
status	complete
benchmark	member-fb.pl
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	BCGGV05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83532 seconds
cpu usage	3.70608
user time	3.53774
system time	0.168344
max virtual memory	1.887396E7
max residence set size	276220.0

stage attributes

key	value
starexec-result	YES

output

3.49/1.79	YES
3.49/1.80	proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
3.49/1.80	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.49/1.80	
3.49/1.80	
3.49/1.80	Left Termination of the query pattern
3.49/1.80	
3.49/1.80	member(a,g)
3.49/1.80	
3.49/1.80	w.r.t. the given Prolog program could successfully be proven:
3.49/1.80	
3.49/1.80	(0) Prolog
3.49/1.80	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.49/1.80	(2) PiTRS
3.49/1.80	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.49/1.80	(4) PiDP
3.49/1.80	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.49/1.80	(6) PiDP
3.49/1.80	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.49/1.80	(8) PiDP
3.49/1.80	(9) PiDPToQDPProof [SOUND, 0 ms]
3.49/1.80	(10) QDP
3.49/1.80	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.49/1.80	(12) YES
3.49/1.80	
3.49/1.80	
3.49/1.80	----------------------------------------
3.49/1.80	
3.49/1.80	(0)
3.49/1.80	Obligation:
3.49/1.80	Clauses:
3.49/1.80	
3.49/1.80	member(X, .(X, X1)).
3.49/1.80	member(X, .(X2, Xs)) :- member(X, Xs).
3.49/1.80	
3.49/1.80	
3.49/1.80	Query: member(a,g)
3.49/1.80	----------------------------------------
3.49/1.80	
3.49/1.80	(1) PrologToPiTRSProof (SOUND)
3.49/1.80	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.49/1.80	
3.49/1.80	member_in_2: (f,b)
3.49/1.80	
3.49/1.80	Transforming Prolog into the following Term Rewriting System:
3.49/1.80	
3.49/1.80	Pi-finite rewrite system:
3.49/1.80	The TRS R consists of the following rules:
3.49/1.80	
3.49/1.80	   member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1))
3.49/1.80	   member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs))
3.49/1.80	   U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs))
3.49/1.80	
3.49/1.80	The argument filtering Pi contains the following mapping:
3.49/1.80	member_in_ag(x1, x2)  =  member_in_ag(x2)
3.49/1.80	
3.49/1.80	.(x1, x2)  =  .(x1, x2)
3.49/1.80	
3.49/1.80	member_out_ag(x1, x2)  =  member_out_ag(x1)
3.49/1.80	
3.49/1.80	U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
3.49/1.80	
3.49/1.80	
3.49/1.80	
3.49/1.80	
3.49/1.80	
3.49/1.80	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.49/1.80	
3.49/1.80	
3.49/1.80	
3.49/1.80	----------------------------------------
3.49/1.80	
3.49/1.80	(2)
3.49/1.80	Obligation:
3.49/1.80	Pi-finite rewrite system:
3.49/1.80	The TRS R consists of the following rules:
3.49/1.80	
3.49/1.80	   member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1))
3.49/1.80	   member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs))
3.49/1.80	   U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs))
3.49/1.80	
3.49/1.80	The argument filtering Pi contains the following mapping:
3.49/1.80	member_in_ag(x1, x2)  =  member_in_ag(x2)
3.49/1.80	
3.49/1.80	.(x1, x2)  =  .(x1, x2)
3.49/1.80	
3.49/1.80	member_out_ag(x1, x2)  =  member_out_ag(x1)
3.49/1.80	
3.49/1.80	U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
3.49/1.80	
3.49/1.80	
3.49/1.80	
3.49/1.80	----------------------------------------
3.49/1.80	
3.49/1.80	(3) DependencyPairsProof (EQUIVALENT)
3.49/1.80	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.49/1.80	Pi DP problem:
3.49/1.80	The TRS P consists of the following rules:
3.49/1.80	
3.49/1.80	   MEMBER_IN_AG(X, .(X2, Xs)) -> U1_AG(X, X2, Xs, member_in_ag(X, Xs))

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