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C Integ Progr 58257 pair #381736501
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
n093.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.71789193153 seconds
cpu usage
7.789508499
max memory
6.43883008E8
stage attributes
key
value
output-size
4497
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 51 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 13 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (8) IntTRS (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (12) IntTRS (13) RankingReductionPairProof [EQUIVALENT, 6 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) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f7(x, x1)] = x [f4(x2, x3)] = x2 The following rules are decreasing:
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