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C Integ Progr 58257 pair #381736873
details
property
value
status
complete
benchmark
LeikeHeizmann-TACAS2014-Ex1_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n006.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.08217310905 seconds
cpu usage
5.383168702
max memory
4.769792E8
stage attributes
key
value
output-size
3339
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 50 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 46 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 14 ms] (8) IntTRS (9) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox2/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(arith, x4) :|: TRUE && arith = x3 - x4 - 1 f6(x21, x22) -> f9(x23, x22) :|: TRUE && x23 = x21 + x22 - 1 f4(x7, x8) -> f5(x7, x8) :|: x8 > 0 f4(x9, x10) -> f6(x9, x10) :|: x10 <= 0 f8(x11, x12) -> f7(x11, x12) :|: TRUE f9(x13, x14) -> f7(x13, x14) :|: TRUE f3(x15, x16) -> f4(x15, x16) :|: x15 > 0 f7(x17, x18) -> f3(x17, x18) :|: TRUE f3(x19, x20) -> f10(x19, x20) :|: x19 <= 0 Start term: f1(q, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x15, x16) -> f4(x15, x16) :|: x15 > 0 f7(x17, x18) -> f3(x17, x18) :|: TRUE f8(x11, x12) -> f7(x11, x12) :|: TRUE f5(x3, x4) -> f8(arith, x4) :|: TRUE && arith = x3 - x4 - 1 f4(x7, x8) -> f5(x7, x8) :|: x8 > 0 f9(x13, x14) -> f7(x13, x14) :|: TRUE f6(x21, x22) -> f9(x23, x22) :|: TRUE && x23 = x21 + x22 - 1 f4(x9, x10) -> f6(x9, x10) :|: x10 <= 0 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1 f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0 ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f7(x, x1)] = -2 + x + x1 The following rules are decreasing: f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1 f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0 The following rules are bounded: f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0 ---------------------------------------- (8) Obligation:
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