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C Integ Progr 58257 pair #381736057
details
property
value
status
complete
benchmark
PastaB14.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n035.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.19134902954 seconds
cpu usage
5.751950973
max memory
4.9348608E8
stage attributes
key
value
output-size
3667
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 49 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 14 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (8) IntTRS (9) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 3 ms] (12) 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) -> f6(arith, x5) :|: TRUE && arith = x4 - 1 f6(x20, x21) -> f7(x20, x22) :|: TRUE && x22 = x21 - 1 f4(x8, x9) -> f5(x8, x9) :|: x9 > 0 f7(x10, x11) -> f4(x10, x11) :|: TRUE f4(x12, x13) -> f8(x12, x13) :|: x13 <= 0 f3(x14, x15) -> f4(x14, x15) :|: x14 = x15 && x14 > 0 f8(x16, x17) -> f3(x16, x17) :|: TRUE f3(x18, x19) -> f9(x18, x19) :|: x18 <= 0 f3(x23, x24) -> f9(x23, x24) :|: x23 < x24 f3(x25, x26) -> f9(x25, x26) :|: x25 > x26 Start term: f1(x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x14, x15) -> f4(x14, x15) :|: x14 = x15 && x14 > 0 f8(x16, x17) -> f3(x16, x17) :|: TRUE f4(x12, x13) -> f8(x12, x13) :|: x13 <= 0 f7(x10, x11) -> f4(x10, x11) :|: TRUE f6(x20, x21) -> f7(x20, x22) :|: TRUE && x22 = x21 - 1 f5(x4, x5) -> f6(arith, x5) :|: TRUE && arith = x4 - 1 f4(x8, x9) -> f5(x8, x9) :|: x9 > 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f4(x8:0, x9:0) -> f4(x8:0 - 1, x9:0 - 1) :|: x9:0 > 0 f4(x12:0, x12:01) -> f4(x12:0, x12:0) :|: x12:0 < 1 && x12:0 > 0 && x12:0 = x12:01 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f4(x, x1)] = d The following rules are decreasing: f4(x12:0, x12:01) -> f4(x12:0, x12:0) :|: x12:0 < 1 && x12:0 > 0 && x12:0 = x12:01 The following rules are bounded: f4(x12:0, x12:01) -> f4(x12:0, x12:0) :|: x12:0 < 1 && x12:0 > 0 && x12:0 = x12:01 ---------------------------------------- (8)
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