details

property	value
status	complete
benchmark	svcomp_c.01_assume.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	3.2148399353 seconds
cpu usage	7.128213268
max memory	6.12806656E8

stage attributes

key	value
output-size	4728
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 80 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 8 ms] (8) IntTRS (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (12) IntTRS (13) PolynomialOrderProcessor [EQUIVALENT, 3 ms] (14) 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) -> f2(x_1, y) :|: TRUE f2(x1, x2) -> f3(x1, x3) :|: TRUE f4(x4, x5) -> f5(x4, 1) :|: TRUE f6(x6, x7) -> f7(x6, arith) :|: TRUE && arith = 2 * x7 f5(x8, x9) -> f6(x8, x9) :|: x8 > x9 && x9 > 0 f7(x10, x11) -> f5(x10, x11) :|: TRUE f5(x12, x13) -> f8(x12, x13) :|: x12 <= x13 f5(x22, x23) -> f8(x22, x23) :|: x23 <= 0 f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 1 f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0 && x17 > 0 f9(x18, x19) -> f3(x18, x19) :|: TRUE f3(x20, x21) -> f10(x20, x21) :|: x20 < 0 f3(x27, x28) -> f10(x27, x28) :|: x28 <= 0 Start term: f1(x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0 && x17 > 0 f9(x18, x19) -> f3(x18, x19) :|: TRUE f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 1 f5(x12, x13) -> f8(x12, x13) :|: x12 <= x13 f4(x4, x5) -> f5(x4, 1) :|: TRUE f7(x10, x11) -> f5(x10, x11) :|: TRUE f6(x6, x7) -> f7(x6, arith) :|: TRUE && arith = 2 * x7 f5(x8, x9) -> f6(x8, x9) :|: x8 > x9 && x9 > 0 f5(x22, x23) -> f8(x22, x23) :|: x23 <= 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x8:0, x9:0) -> f5(x8:0, 2 * x9:0) :|: x9:0 < x8:0 && x9:0 > 0 f5(x12:0, x13:0) -> f5(x12:0 - 1, 1) :|: x12:0 > 0 && x13:0 > 0 && x13:0 >= x12:0 f5(x22:0, x23:0) -> f5(x22:0 - 1, 1) :|: x22:0 > 0 && x23:0 > 0 && x23:0 < 1 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f5(x, x1)] = d The following rules are decreasing: f5(x22:0, x23:0) -> f5(x22:0 - 1, 1) :|: x22:0 > 0 && x23:0 > 0 && x23:0 < 1 popout

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job log

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actions

all output return to C Integer