details

property	value
status	complete
benchmark	Nyala-2lex_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n013.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.3025329113 seconds
cpu usage	6.355390024
max memory	5.147648E8

stage attributes

key	value
output-size	3325
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 46 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 26 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 12 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox2/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1 f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1 f7(x3, x4) -> f8(x3, x5) :|: TRUE f3(x6, x7) -> f4(x6, x7) :|: x7 < 0 f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0 f8(x10, x11) -> f6(x10, x11) :|: TRUE f5(x12, x13) -> f6(x12, x13) :|: TRUE f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0 f6(x16, x17) -> f1(x16, x17) :|: TRUE f1(x18, x19) -> f9(x18, x19) :|: x18 < 0 f1(x23, x24) -> f9(x23, x24) :|: x24 < 0 Start term: f1(x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0 f6(x16, x17) -> f1(x16, x17) :|: TRUE f8(x10, x11) -> f6(x10, x11) :|: TRUE f7(x3, x4) -> f8(x3, x5) :|: TRUE f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1 f3(x6, x7) -> f4(x6, x7) :|: x7 < 0 f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1 f5(x12, x13) -> f6(x12, x13) :|: TRUE f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f3(x8:0, x9:0) -> f3(x8:0, x9:0 - 1) :|: x9:0 > -1 && x8:0 > -1 f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f3(x, x1)] = -1 + x The following rules are decreasing: f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0 The following rules are bounded: f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0 ---------------------------------------- (8) Obligation: popout

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

job log

popout

actions

all output return to C Integ Progr 58257