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C_Integer 2019-03-28 22.15 pair #432271957
details
property
value
status
complete
benchmark
ChenFlurMukhopadhyay-SAS2012-Ex3.10_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n122.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.18314 seconds
cpu usage
5.48667
user time
5.24626
system time
0.240411
max virtual memory
1.907664E7
max residence set size
419012.0
stage attributes
key
value
starexec-result
YES
output
5.17/2.13 YES 5.17/2.14 proof of /export/starexec/sandbox/benchmark/theBenchmark.c 5.17/2.14 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 5.17/2.14 5.17/2.14 5.17/2.14 Termination of the given C Problem could be proven: 5.17/2.14 5.17/2.14 (0) C Problem 5.17/2.14 (1) CToIRSProof [EQUIVALENT, 0 ms] 5.17/2.14 (2) IntTRS 5.17/2.14 (3) TerminationGraphProcessor [SOUND, 48 ms] 5.17/2.14 (4) IntTRS 5.17/2.14 (5) IntTRSCompressionProof [EQUIVALENT, 40 ms] 5.17/2.14 (6) IntTRS 5.17/2.14 (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] 5.17/2.14 (8) IntTRS 5.17/2.14 (9) PolynomialOrderProcessor [EQUIVALENT, 7 ms] 5.17/2.14 (10) YES 5.17/2.14 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (0) 5.17/2.14 Obligation: 5.17/2.14 c file /export/starexec/sandbox/benchmark/theBenchmark.c 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (1) CToIRSProof (EQUIVALENT) 5.17/2.14 Parsed C Integer Program as IRS. 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (2) 5.17/2.14 Obligation: 5.17/2.14 Rules: 5.17/2.14 f1(x, y, z) -> f2(x_1, y, z) :|: TRUE 5.17/2.14 f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE 5.17/2.14 f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE 5.17/2.14 f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10 + x11 5.17/2.14 f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 0 - x26 - 1 5.17/2.14 f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 >= 0 && x15 + x16 >= 0 5.17/2.14 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE 5.17/2.14 f4(x21, x22, x23) -> f8(x21, x22, x23) :|: x21 < 0 5.17/2.14 f4(x28, x29, x30) -> f8(x28, x29, x30) :|: x28 + x29 < 0 5.17/2.14 Start term: f1(x, y, z) 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (3) TerminationGraphProcessor (SOUND) 5.17/2.14 Constructed the termination graph and obtained one non-trivial SCC. 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (4) 5.17/2.14 Obligation: 5.17/2.14 Rules: 5.17/2.14 f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 >= 0 && x15 + x16 >= 0 5.17/2.14 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE 5.17/2.14 f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 0 - x26 - 1 5.17/2.14 f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10 + x11 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (5) IntTRSCompressionProof (EQUIVALENT) 5.17/2.14 Compressed rules. 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (6) 5.17/2.14 Obligation: 5.17/2.14 Rules: 5.17/2.14 f6(x24:0, x25:0, x26:0) -> f6(x24:0 + (0 - x26:0 - 1) + x26:0, 0 - x26:0 - 1, x26:0) :|: x24:0 > -1 && x24:0 + (0 - x26:0 - 1) >= 0 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (7) IntTRSUnneededArgumentFilterProof (EQUIVALENT) 5.17/2.14 Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: 5.17/2.14 5.17/2.14 f6(x1, x2, x3) -> f6(x1, x3) 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (8) 5.17/2.14 Obligation: 5.17/2.14 Rules: 5.17/2.14 f6(x24:0, x26:0) -> f6(x24:0 + (0 - x26:0 - 1) + x26:0, x26:0) :|: x24:0 > -1 && x24:0 + (0 - x26:0 - 1) >= 0 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (9) PolynomialOrderProcessor (EQUIVALENT) 5.17/2.14 Found the following polynomial interpretation: 5.17/2.14 [f6(x, x1)] = x 5.17/2.14 5.17/2.14 The following rules are decreasing: 5.17/2.14 f6(x24:0, x26:0) -> f6(x24:0 + (0 - x26:0 - 1) + x26:0, x26:0) :|: x24:0 > -1 && x24:0 + (0 - x26:0 - 1) >= 0 5.17/2.14 The following rules are bounded: 5.17/2.14 f6(x24:0, x26:0) -> f6(x24:0 + (0 - x26:0 - 1) + x26:0, x26:0) :|: x24:0 > -1 && x24:0 + (0 - x26:0 - 1) >= 0 5.17/2.14 5.17/2.14 ---------------------------------------- 5.17/2.14 5.17/2.14 (10) 5.17/2.14 YES
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