details

property	value
status	complete
benchmark	svcomp_a.05.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.1821501255 seconds
cpu usage	5.372822948
max memory	4.1064448E8

stage attributes

key	value
output-size	2957
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 51 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, x9, arith) :|: TRUE && arith = x10 + 1 f6(x23, x24, x25) -> f7(x26, x24, x25) :|: TRUE && x26 = x23 + 1 f4(x14, x15, x16) -> f5(x14, x15, x16) :|: x15 >= x16 + 1 f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE f4(x20, x21, x22) -> f8(x20, x21, x22) :|: x21 < x22 + 1 Start term: f1(c, x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f4(x14, x15, x16) -> f5(x14, x15, x16) :|: x15 >= x16 + 1 f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE f6(x23, x24, x25) -> f7(x26, x24, x25) :|: TRUE && x26 = x23 + 1 f5(x8, x9, x10) -> f6(x8, x9, arith) :|: TRUE && arith = x10 + 1 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f6(x23:0, x24:0, x25:0) -> f6(x23:0 + 1, x24:0, x25:0 + 1) :|: x25:0 + 1 <= x24:0 ---------------------------------------- (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(x2, x3) ---------------------------------------- (8) Obligation: Rules: f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 + 1 <= x24:0 ---------------------------------------- (9) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f6(x, x1)] = x - x1 The following rules are decreasing: f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 + 1 <= x24:0 The following rules are bounded: f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 + 1 <= x24:0 popout

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actions

all output return to C Integ Progr 58257