details

property	value
status	complete
benchmark	sum.pl
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n046.star.cs.uiowa.edu
space	talp_apt

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.81405 seconds
cpu usage	4.03762
user time	3.84944
system time	0.188187
max virtual memory	1.8877384E7
max residence set size	273532.0

stage attributes

key	value
starexec-result	YES

output

3.87/1.74 YES 3.87/1.75 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.87/1.75 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.87/1.75 3.87/1.75 3.87/1.75 Left Termination of the query pattern 3.87/1.75 3.87/1.75 sum(a,a,g) 3.87/1.75 3.87/1.75 w.r.t. the given Prolog program could successfully be proven: 3.87/1.75 3.87/1.75 (0) Prolog 3.87/1.75 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.87/1.75 (2) PiTRS 3.87/1.75 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.87/1.75 (4) PiDP 3.87/1.75 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.87/1.75 (6) PiDP 3.87/1.75 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.87/1.75 (8) PiDP 3.87/1.75 (9) PiDPToQDPProof [SOUND, 0 ms] 3.87/1.75 (10) QDP 3.87/1.75 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.87/1.75 (12) YES 3.87/1.75 3.87/1.75 3.87/1.75 ---------------------------------------- 3.87/1.75 3.87/1.75 (0) 3.87/1.75 Obligation: 3.87/1.75 Clauses: 3.87/1.75 3.87/1.75 sum(X, s(Y), s(Z)) :- sum(X, Y, Z). 3.87/1.75 sum(X, 0, X). 3.87/1.75 3.87/1.75 3.87/1.75 Query: sum(a,a,g) 3.87/1.75 ---------------------------------------- 3.87/1.75 3.87/1.75 (1) PrologToPiTRSProof (SOUND) 3.87/1.75 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.87/1.75 3.87/1.75 sum_in_3: (f,f,b) 3.87/1.75 3.87/1.75 Transforming Prolog into the following Term Rewriting System: 3.87/1.75 3.87/1.75 Pi-finite rewrite system: 3.87/1.75 The TRS R consists of the following rules: 3.87/1.75 3.87/1.75 sum_in_aag(X, s(Y), s(Z)) -> U1_aag(X, Y, Z, sum_in_aag(X, Y, Z)) 3.87/1.75 sum_in_aag(X, 0, X) -> sum_out_aag(X, 0, X) 3.87/1.75 U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) -> sum_out_aag(X, s(Y), s(Z)) 3.87/1.75 3.87/1.75 The argument filtering Pi contains the following mapping: 3.87/1.75 sum_in_aag(x1, x2, x3) = sum_in_aag(x3) 3.87/1.75 3.87/1.75 s(x1) = s(x1) 3.87/1.75 3.87/1.75 U1_aag(x1, x2, x3, x4) = U1_aag(x3, x4) 3.87/1.75 3.87/1.75 sum_out_aag(x1, x2, x3) = sum_out_aag(x1, x2, x3) 3.87/1.75 3.87/1.75 3.87/1.75 3.87/1.75 3.87/1.75 3.87/1.75 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.87/1.75 3.87/1.75 3.87/1.75 3.87/1.75 ---------------------------------------- 3.87/1.75 3.87/1.75 (2) 3.87/1.75 Obligation: 3.87/1.75 Pi-finite rewrite system: 3.87/1.75 The TRS R consists of the following rules: 3.87/1.75 3.87/1.75 sum_in_aag(X, s(Y), s(Z)) -> U1_aag(X, Y, Z, sum_in_aag(X, Y, Z)) 3.87/1.75 sum_in_aag(X, 0, X) -> sum_out_aag(X, 0, X) 3.87/1.75 U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) -> sum_out_aag(X, s(Y), s(Z)) 3.87/1.75 3.87/1.75 The argument filtering Pi contains the following mapping: 3.87/1.75 sum_in_aag(x1, x2, x3) = sum_in_aag(x3) 3.87/1.75 3.87/1.75 s(x1) = s(x1) 3.87/1.75 3.87/1.75 U1_aag(x1, x2, x3, x4) = U1_aag(x3, x4) 3.87/1.75 3.87/1.75 sum_out_aag(x1, x2, x3) = sum_out_aag(x1, x2, x3) 3.87/1.75 3.87/1.75 3.87/1.75 3.87/1.75 ---------------------------------------- 3.87/1.75 3.87/1.75 (3) DependencyPairsProof (EQUIVALENT) 3.87/1.75 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.87/1.75 Pi DP problem: 3.87/1.75 The TRS P consists of the following rules: 3.87/1.75 3.87/1.75 SUM_IN_AAG(X, s(Y), s(Z)) -> U1_AAG(X, Y, Z, sum_in_aag(X, Y, Z)) popout

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