details

property	value
status	complete
benchmark	AliasDarteFeautrierGonnord-SAS2010-wcet2_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	2.64498496056 seconds
cpu usage	7.075588781
max memory	5.86285056E8

stage attributes

key	value
output-size	4051
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 68 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 15 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox2/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(i, j) -> f2(x_1, j) :|: TRUE f2(x, x1) -> f3(x, x2) :|: TRUE f4(x3, x4) -> f5(x3, 0) :|: TRUE f6(x5, x6) -> f7(x5, arith) :|: TRUE && arith = x6 + 1 f5(x7, x8) -> f6(x7, x8) :|: x7 > 2 && x8 <= 9 f7(x9, x10) -> f5(x9, x10) :|: TRUE f5(x11, x12) -> f8(x11, x12) :|: x11 <= 2 f5(x21, x22) -> f8(x21, x22) :|: x22 > 9 f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f3(x15, x16) -> f4(x15, x16) :|: x15 < 5 f9(x17, x18) -> f3(x17, x18) :|: TRUE f3(x19, x20) -> f10(x19, x20) :|: x19 >= 5 Start term: f1(i, j) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x15, x16) -> f4(x15, x16) :|: x15 < 5 f9(x17, x18) -> f3(x17, x18) :|: TRUE f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f5(x11, x12) -> f8(x11, x12) :|: x11 <= 2 f4(x3, x4) -> f5(x3, 0) :|: TRUE f7(x9, x10) -> f5(x9, x10) :|: TRUE f6(x5, x6) -> f7(x5, arith) :|: TRUE && arith = x6 + 1 f5(x7, x8) -> f6(x7, x8) :|: x7 > 2 && x8 <= 9 f5(x21, x22) -> f8(x21, x22) :|: x22 > 9 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x7:0, x8:0) -> f5(x7:0, x8:0 + 1) :|: x7:0 > 2 && x8:0 < 10 f5(x11:0, x12:0) -> f5(x11:0 + 1, 0) :|: x11:0 < 3 && x11:0 < 4 f5(x21:0, x22:0) -> f5(x21:0 + 1, 0) :|: x22:0 > 9 && x21:0 < 4 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f5(x, x1)] = 2 - x The following rules are decreasing: f5(x11:0, x12:0) -> f5(x11:0 + 1, 0) :|: x11:0 < 3 && x11:0 < 4 f5(x21:0, x22:0) -> f5(x21:0 + 1, 0) :|: x22:0 > 9 && x21:0 < 4 The following rules are bounded: f5(x11:0, x12:0) -> f5(x11:0 + 1, 0) :|: x11:0 < 3 && x11:0 < 4 popout

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job log

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actions

all output return to C Integer