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C_Integer pair #487095237
details
property
value
status
complete
benchmark
Waldkirch_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n138.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.00242 seconds
cpu usage
4.87437
user time
4.6476
system time
0.226768
max virtual memory
1.8946796E7
max residence set size
375864.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.c # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 24 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 9 ms] (6) IntTRS (7) RankingReductionPairProof [EQUIVALENT, 0 ms] (8) 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) -> f2(0) :|: TRUE f3(x1) -> f4(arith) :|: TRUE && arith = x1 - 1 f2(x2) -> f3(x2) :|: x2 >= 0 f4(x3) -> f2(x3) :|: TRUE f2(x4) -> f5(x4) :|: x4 < 0 Start term: f1(x) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f2(x2) -> f3(x2) :|: x2 >= 0 f4(x3) -> f2(x3) :|: TRUE f3(x1) -> f4(arith) :|: TRUE && arith = x1 - 1 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f4(x3:0) -> f4(x3:0 - 1) :|: x3:0 > -1 ---------------------------------------- (7) RankingReductionPairProof (EQUIVALENT) Interpretation: [ f4 ] = f4_1 The following rules are decreasing: f4(x3:0) -> f4(x3:0 - 1) :|: x3:0 > -1 The following rules are bounded: f4(x3:0) -> f4(x3:0 - 1) :|: x3:0 > -1 ---------------------------------------- (8) YES
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