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C Integ Progr 58257 pair #381736714
details
property
value
status
complete
benchmark
ChenFlurMukhopadhyay-SAS2012-Ex1.03_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n055.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
1.91270399094 seconds
cpu usage
4.884020231
max memory
3.95010048E8
stage attributes
key
value
output-size
2714
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_c /export/starexec/sandbox2/benchmark/theBenchmark.c /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.c # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 50 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 11 ms] (10) 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, oldx) -> f2(x_1, oldx) :|: TRUE f3(x1, x2) -> f4(x1, x1) :|: TRUE f4(x3, x4) -> f5(x5, x4) :|: TRUE f2(x6, x7) -> f3(x6, x7) :|: x6 > 1 && 0 - 2 * x6 = x7 f5(x8, x9) -> f2(x8, x9) :|: TRUE f2(x10, x11) -> f6(x10, x11) :|: x10 <= 1 f2(x12, x13) -> f6(x12, x13) :|: 0 - 2 * x12 < x13 f2(x14, x15) -> f6(x14, x15) :|: 0 - 2 * x14 > x15 Start term: f1(x, oldx) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f2(x6, x7) -> f3(x6, x7) :|: x6 > 1 && 0 - 2 * x6 = x7 f5(x8, x9) -> f2(x8, x9) :|: TRUE f4(x3, x4) -> f5(x5, x4) :|: TRUE f3(x1, x2) -> f4(x1, x1) :|: TRUE ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f4(x, x1) -> f4(x2, x2) :|: x2 > 1 && x1 = 0 - 2 * x2 ---------------------------------------- (7) IntTRSUnneededArgumentFilterProof (EQUIVALENT) Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: f4(x1, x2) -> f4(x2) ---------------------------------------- (8) Obligation: Rules: f4(x1) -> f4(x2) :|: x2 > 1 && x1 = 0 - 2 * x2 ---------------------------------------- (9) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f4(x)] = -x The following rules are decreasing: f4(x1) -> f4(x2) :|: x2 > 1 && x1 = 0 - 2 * x2 The following rules are bounded: f4(x1) -> f4(x2) :|: x2 > 1 && x1 = 0 - 2 * x2
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