details

property	value
status	complete
benchmark	Parallel_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.66405 seconds
cpu usage	7.44226
user time	7.02695
system time	0.415311
max virtual memory	1.9401584E7
max residence set size	600776.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.c # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 66 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 9 ms] (6) IntTRS (7) RankingReductionPairProof [EQUIVALENT, 37 ms] (8) IntTRS (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] (10) IntTRS (11) RankingReductionPairProof [EQUIVALENT, 8 ms] (12) IntTRS (13) PolynomialOrderProcessor [EQUIVALENT, 1 ms] (14) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox2/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(x, y) -> f2(x_1, y) :|: TRUE f2(x1, x2) -> f3(x1, x3) :|: TRUE f5(x4, x5) -> f8(arith, x5) :|: TRUE && arith = x4 - 1 f6(x22, x23) -> f9(x22, x24) :|: TRUE && x24 = x23 - 1 f4(x8, x9) -> f5(x8, x9) :|: x8 >= 0 f4(x10, x11) -> f6(x10, x11) :|: x10 < 0 f8(x12, x13) -> f7(x12, x13) :|: TRUE f9(x14, x15) -> f7(x14, x15) :|: TRUE f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0 f3(x25, x26) -> f4(x25, x26) :|: x26 >= 0 f7(x18, x19) -> f3(x18, x19) :|: TRUE f3(x20, x21) -> f10(x20, x21) :|: x20 < 0 && x21 < 0 Start term: f1(x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0 f7(x18, x19) -> f3(x18, x19) :|: TRUE f8(x12, x13) -> f7(x12, x13) :|: TRUE f5(x4, x5) -> f8(arith, x5) :|: TRUE && arith = x4 - 1 f4(x8, x9) -> f5(x8, x9) :|: x8 >= 0 f3(x25, x26) -> f4(x25, x26) :|: x26 >= 0 f9(x14, x15) -> f7(x14, x15) :|: TRUE f6(x22, x23) -> f9(x22, x24) :|: TRUE && x24 = x23 - 1 f4(x10, x11) -> f6(x10, x11) :|: x10 < 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f7(x18:0, x19:0) -> f4(x18:0, x19:0) :|: x19:0 > -1 f4(x10:0, x11:0) -> f7(x10:0, x11:0 - 1) :|: x10:0 < 0 f7(x, x1) -> f4(x, x1) :|: x > -1 f4(x8:0, x9:0) -> f7(x8:0 - 1, x9:0) :|: x8:0 > -1 ---------------------------------------- (7) RankingReductionPairProof (EQUIVALENT) Interpretation: [ f7 ] = 3*f7_1 [ f4 ] = 3*f4_1 The following rules are decreasing: f4(x8:0, x9:0) -> f7(x8:0 - 1, x9:0) :|: x8:0 > -1 The following rules are bounded: f4(x8:0, x9:0) -> f7(x8:0 - 1, x9:0) :|: x8:0 > -1 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to C_Integer