details

property	value
status	complete
benchmark	Urban-WST2013-Fig2-modified1000_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.16554 seconds
cpu usage	5.7217
user time	5.37596
system time	0.345742
max virtual memory	1.9210908E7
max residence set size	478576.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 43 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 11 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 13 ms] (8) AND (9) IntTRS (10) TerminationGraphProcessor [EQUIVALENT, 0 ms] (11) YES (12) IntTRS (13) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (14) IntTRS (15) PolynomialOrderProcessor [EQUIVALENT, 3 ms] (16) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(x1, x2) -> f2(x_1, x2) :|: TRUE f2(x, x3) -> f3(x, x4) :|: TRUE f4(x5, x6) -> f5(x5, 1000) :|: TRUE f6(x7, x8) -> f7(x7, arith) :|: TRUE && arith = x8 - 1 f5(x9, x10) -> f6(x9, x10) :|: x10 > 1 f7(x11, x12) -> f5(x11, x12) :|: TRUE f5(x13, x14) -> f8(x13, x14) :|: x14 <= 1 f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f3(x17, x18) -> f4(x17, x18) :|: x17 <= 10 f9(x19, x20) -> f3(x19, x20) :|: TRUE f3(x21, x22) -> f10(x21, x22) :|: x21 > 10 Start term: f1(x1, x2) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x17, x18) -> f4(x17, x18) :|: x17 <= 10 f9(x19, x20) -> f3(x19, x20) :|: TRUE f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f5(x13, x14) -> f8(x13, x14) :|: x14 <= 1 f7(x11, x12) -> f5(x11, x12) :|: TRUE f6(x7, x8) -> f7(x7, arith) :|: TRUE && arith = x8 - 1 f5(x9, x10) -> f6(x9, x10) :|: x10 > 1 f4(x5, x6) -> f5(x5, 1000) :|: TRUE ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x13:0, x14:0) -> f5(x13:0 + 1, 1000) :|: x14:0 < 2 && x13:0 < 10 f5(x9:0, x10:0) -> f5(x9:0, x10:0 - 1) :|: x10:0 > 1 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f5(x, x1)] = 8982010 - 998001*x - 2*x1 + x1^2 The following rules are decreasing: f5(x9:0, x10:0) -> f5(x9:0, x10:0 - 1) :|: x10:0 > 1 The following rules are bounded: f5(x13:0, x14:0) -> f5(x13:0 + 1, 1000) :|: x14:0 < 2 && x13:0 < 10 ---------------------------------------- (8) Complex Obligation (AND) popout

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job log

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actions

all output return to C_Integer