details

property	value
status	complete
benchmark	Copenhagen_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n007.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.1342959404 seconds
cpu usage	5.437940734
max memory	4.04533248E8

stage attributes

key	value
output-size	3066
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 55 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 19 ms] (6) IntTRS (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (8) IntTRS (9) RankingReductionPairProof [EQUIVALENT, 0 ms] (10) 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, y, oldx) -> f2(x_1, y, oldx) :|: TRUE f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1 f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1 f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0 && x15 >= 0 f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE f3(x20, x21, x22) -> f8(x20, x21, x22) :|: x20 < 0 f3(x27, x28, x29) -> f8(x27, x28, x29) :|: x28 < 0 Start term: f1(x, y, oldx) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0 && x15 >= 0 f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1 f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1 f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x8:0, x9:0, x10:0) -> f5(x9:0 - 1, x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0 ---------------------------------------- (7) IntTRSUnneededArgumentFilterProof (EQUIVALENT) Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: f5(x1, x2, x3) -> f5(x2, x3) ---------------------------------------- (8) Obligation: Rules: f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0 ---------------------------------------- (9) RankingReductionPairProof (EQUIVALENT) Interpretation: [ f5 ] = 1/2*f5_1 + 1/2*f5_2 The following rules are decreasing: f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0 popout

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actions

all output return to C Integ Progr 58257