details

property	value
status	complete
benchmark	svcomp_b.06.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n099.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	1.92323994637 seconds
cpu usage	4.829348092
max memory	4.08731648E8

stage attributes

key	value
output-size	3170
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 50 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 4 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(c, x, y) -> f2(c, x_1, y) :|: TRUE f2(x1, x2, x3) -> f3(x1, x2, x4) :|: TRUE f3(x5, x6, x7) -> f4(0, x6, x7) :|: TRUE f5(x8, x9, x10) -> f6(x8, arith, x10) :|: TRUE && arith = x9 - 1 f6(x26, x27, x28) -> f7(x26, x27, x29) :|: TRUE && x29 = x28 - 1 f7(x30, x31, x32) -> f8(x33, x31, x32) :|: TRUE && x33 = x30 + 1 f4(x17, x18, x19) -> f5(x17, x18, x19) :|: x18 > 0 && x19 > 0 f8(x20, x21, x22) -> f4(x20, x21, x22) :|: TRUE f4(x23, x24, x25) -> f9(x23, x24, x25) :|: x24 <= 0 f4(x34, x35, x36) -> f9(x34, x35, x36) :|: x36 <= 0 Start term: f1(c, x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f4(x17, x18, x19) -> f5(x17, x18, x19) :|: x18 > 0 && x19 > 0 f8(x20, x21, x22) -> f4(x20, x21, x22) :|: TRUE f7(x30, x31, x32) -> f8(x33, x31, x32) :|: TRUE && x33 = x30 + 1 f6(x26, x27, x28) -> f7(x26, x27, x29) :|: TRUE && x29 = x28 - 1 f5(x8, x9, x10) -> f6(x8, arith, x10) :|: TRUE && arith = x9 - 1 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f6(x26:0, x27:0, x28:0) -> f6(x26:0 + 1, x27:0 - 1, x28:0 - 1) :|: x27:0 > 0 && x28:0 > 1 ---------------------------------------- (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(x2, x3) ---------------------------------------- (8) Obligation: Rules: f6(x27:0, x28:0) -> f6(x27:0 - 1, x28:0 - 1) :|: x27:0 > 0 && x28:0 > 1 ---------------------------------------- (9) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f6(x, x1)] = x1 The following rules are decreasing: popout

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

job log

popout

actions

all output return to C Integer