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C Integ Progr 58257 pair #381736432
details
property
value
status
complete
benchmark
ChenFlurMukhopadhyay-SAS2012-Ex2.09_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n028.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.63753890991 seconds
cpu usage
7.689883669
max memory
6.54508032E8
stage attributes
key
value
output-size
5038
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 46 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) TerminationGraphProcessor [EQUIVALENT, 4 ms] (8) IntTRS (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] (10) IntTRS (11) RankingReductionPairProof [EQUIVALENT, 11 ms] (12) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(x, y, n) -> f2(x_1, y, n) :|: TRUE f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = 0 - x9 + x10 - 5 f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 2 * x25 f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 > 0 && x15 < x17 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE f4(x21, x22, x23) -> f8(x21, x22, x23) :|: x21 <= 0 f4(x28, x29, x30) -> f8(x28, x29, x30) :|: x28 >= x30 Start term: f1(x, y, n) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 > 0 && x15 < x17 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 2 * x25 f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = 0 - x9 + x10 - 5 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f6(x24:0, x25:0, x26:0) -> f6(0 - x24:0 + 2 * x25:0 - 5, 2 * x25:0, x26:0) :|: x24:0 > 0 && x26:0 > x24:0 ---------------------------------------- (7) TerminationGraphProcessor (EQUIVALENT) Constructed the termination graph and obtained one non-trivial SCC. f6(x24:0, x25:0, x26:0) -> f6(0 - x24:0 + 2 * x25:0 - 5, 2 * x25:0, x26:0) :|: x24:0 > 0 && x26:0 > x24:0 has been transformed into f6(x24:0, x25:0, x26:0) -> f6(0 - x24:0 + 2 * x25:0 - 5, 2 * x25:0, x26:0) :|: x26:0 = x8 && (x24:0 > 0 && x26:0 > x24:0) && x6 > 0 && x8 > x6. f6(x24:0, x25:0, x26:0) -> f6(0 - x24:0 + 2 * x25:0 - 5, 2 * x25:0, x26:0) :|: x26:0 = x8 && (x24:0 > 0 && x26:0 > x24:0) && x6 > 0 && x8 > x6 and f6(x24:0, x25:0, x26:0) -> f6(0 - x24:0 + 2 * x25:0 - 5, 2 * x25:0, x26:0) :|: x26:0 = x8 && (x24:0 > 0 && x26:0 > x24:0) && x6 > 0 && x8 > x6 have been merged into the new rule f6(x19, x20, x21) -> f6(0 - (0 - x19 + 2 * x20 - 5) + 2 * (2 * x20) - 5, 2 * (2 * x20), x21) :|: x21 = x22 && (x19 > 0 && x21 > x19) && x23 > 0 && x22 > x23 && (x21 = x24 && (0 - x19 + 2 * x20 - 5 > 0 && x21 > 0 - x19 + 2 * x20 - 5) && x25 > 0 && x24 > x25) ---------------------------------------- (8) Obligation: Rules:
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