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C Integ Progr 58257 pair #381736744
details
property
value
status
complete
benchmark
Nested.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n003.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.1454679966 seconds
cpu usage
5.78646154
max memory
4.8111616E8
stage attributes
key
value
output-size
3242
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 51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, 0 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 3 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(i, j) -> f2(0, j) :|: TRUE f2(x, x1) -> f3(x, 3) :|: TRUE f5(x2, x3) -> f6(x2, arith) :|: TRUE && arith = x3 - 1 f6(x20, x21) -> f7(x20, x22) :|: TRUE && x22 = x21 + 2 f4(x6, x7) -> f5(x6, x7) :|: x7 < 12 f7(x8, x9) -> f4(x8, x9) :|: TRUE f4(x10, x11) -> f8(x10, x11) :|: x11 >= 12 f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f3(x14, x15) -> f4(x14, x15) :|: x14 < 10 f9(x16, x17) -> f3(x16, x17) :|: TRUE f3(x18, x19) -> f10(x18, x19) :|: x18 >= 10 Start term: f1(i, j) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x14, x15) -> f4(x14, x15) :|: x14 < 10 f9(x16, x17) -> f3(x16, x17) :|: TRUE f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f4(x10, x11) -> f8(x10, x11) :|: x11 >= 12 f7(x8, x9) -> f4(x8, x9) :|: TRUE f6(x20, x21) -> f7(x20, x22) :|: TRUE && x22 = x21 + 2 f5(x2, x3) -> f6(x2, arith) :|: TRUE && arith = x3 - 1 f4(x6, x7) -> f5(x6, x7) :|: x7 < 12 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f4(x10:0, x11:0) -> f4(x10:0 + 1, x11:0) :|: x11:0 > 11 && x10:0 < 9 f4(x6:0, x7:0) -> f4(x6:0, x7:0 + 1) :|: x7:0 < 12 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f4(x, x1)] = 11 - x1 The following rules are decreasing: f4(x6:0, x7:0) -> f4(x6:0, x7:0 + 1) :|: x7:0 < 12 The following rules are bounded: f4(x6:0, x7:0) -> f4(x6:0, x7:0 + 1) :|: x7:0 < 12 ---------------------------------------- (8) Obligation: Rules:
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