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	n030.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.05246400833 seconds
cpu usage	5.137503282
max memory	2.78134784E8

stage attributes

key	value
output-size	3245
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 43 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 30 ms] (6) IntTRS (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 5 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, z) -> f2(x_1, y, z) :|: TRUE f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10 + x11 f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 0 - x26 - 1 f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 >= 0 && x15 + x16 >= 0 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE f4(x21, x22, x23) -> f8(x21, x22, x23) :|: x21 < 0 f4(x28, x29, x30) -> f8(x28, x29, x30) :|: x28 + x29 < 0 Start term: f1(x, y, z) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f4(x15, x16, x17) -> f5(x15, x16, x17) :|: x15 >= 0 && x15 + x16 >= 0 f7(x18, x19, x20) -> f4(x18, x19, x20) :|: TRUE f6(x24, x25, x26) -> f7(x24, x27, x26) :|: TRUE && x27 = 0 - x26 - 1 f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10 + x11 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: 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 ---------------------------------------- (7) IntTRSUnneededArgumentFilterProof (EQUIVALENT) Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: f6(x1, x2, x3) -> f6(x1, x3) ---------------------------------------- (8) Obligation: Rules: 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 ---------------------------------------- (9) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f6(x, x1)] = x The following rules are decreasing: 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 The following rules are bounded: popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to C Integ Progr 58257