details

property	value
status	complete
benchmark	example4.pl
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	talp_talp

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.8402 seconds
cpu usage	3.83488
user time	3.66322
system time	0.171665
max virtual memory	1.8478712E7
max residence set size	283296.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern p1(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) PiDP (9) PiDPToQDPProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Clauses: p1(f(X)) :- p1(X). p2(f(X)) :- p2(X). Query: p1(g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: p1_in_1: (b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p1_in_g(f(X)) -> U1_g(X, p1_in_g(X)) U1_g(X, p1_out_g(X)) -> p1_out_g(f(X)) The argument filtering Pi contains the following mapping: p1_in_g(x1) = p1_in_g(x1) f(x1) = f(x1) U1_g(x1, x2) = U1_g(x2) p1_out_g(x1) = p1_out_g Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p1_in_g(f(X)) -> U1_g(X, p1_in_g(X)) U1_g(X, p1_out_g(X)) -> p1_out_g(f(X)) The argument filtering Pi contains the following mapping: p1_in_g(x1) = p1_in_g(x1) f(x1) = f(x1) U1_g(x1, x2) = U1_g(x2) p1_out_g(x1) = p1_out_g ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: P1_IN_G(f(X)) -> U1_G(X, p1_in_g(X)) P1_IN_G(f(X)) -> P1_IN_G(X) popout

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