details

property	value
status	complete
benchmark	LeikeHeizmann-TACAS2014-Ex8_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.50886893272 seconds
cpu usage	6.975893736
max memory	3.30387456E8

stage attributes

key	value
output-size	3415
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 59 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (8) IntTRS (9) PolynomialOrderProcessor [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(q, y) -> f2(x_1, y) :|: TRUE f2(x, x1) -> f3(x, x2) :|: TRUE f5(x3, x4) -> f8(x3, 0) :|: TRUE f8(x5, x6) -> f9(x7, x6) :|: TRUE f6(x8, x9) -> f10(x8, arith) :|: TRUE && arith = x9 - 1 f10(x26, x27) -> f11(x28, x27) :|: TRUE && x28 = x26 - 1 f4(x12, x13) -> f5(x12, x13) :|: x13 > 0 f4(x14, x15) -> f6(x14, x15) :|: x15 <= 0 f9(x16, x17) -> f7(x16, x17) :|: TRUE f11(x18, x19) -> f7(x18, x19) :|: TRUE f3(x20, x21) -> f4(x20, x21) :|: x20 > 0 f7(x22, x23) -> f3(x22, x23) :|: TRUE f3(x24, x25) -> f12(x24, x25) :|: x24 <= 0 Start term: f1(q, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x20, x21) -> f4(x20, x21) :|: x20 > 0 f7(x22, x23) -> f3(x22, x23) :|: TRUE f9(x16, x17) -> f7(x16, x17) :|: TRUE f8(x5, x6) -> f9(x7, x6) :|: TRUE f5(x3, x4) -> f8(x3, 0) :|: TRUE f4(x12, x13) -> f5(x12, x13) :|: x13 > 0 f11(x18, x19) -> f7(x18, x19) :|: TRUE f10(x26, x27) -> f11(x28, x27) :|: TRUE && x28 = x26 - 1 f6(x8, x9) -> f10(x8, arith) :|: TRUE && arith = x9 - 1 f4(x14, x15) -> f6(x14, x15) :|: x15 <= 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f7(x22:0, x23:0) -> f7(x22:0 - 1, x23:0 - 1) :|: x22:0 > 0 && x23:0 < 1 f7(x, x1) -> f7(x2, 0) :|: x > 0 && x1 > 0 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f7(x, x1)] = -1 + x1 The following rules are decreasing: f7(x22:0, x23:0) -> f7(x22:0 - 1, x23:0 - 1) :|: x22:0 > 0 && x23:0 < 1 f7(x, x1) -> f7(x2, 0) :|: x > 0 && x1 > 0 The following rules are bounded: f7(x, x1) -> f7(x2, 0) :|: x > 0 && x1 > 0 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