/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 48 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 15 ms] (6) IntTRS (7) RankingReductionPairProof [EQUIVALENT, 27 ms] (8) IntTRS (9) IntTRSUnneededArgumentFilterProof [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(x1, x2) -> f2(x_1, x2) :|: TRUE f2(x, x3) -> f3(x, x4) :|: TRUE f4(x5, x6) -> f5(x5, 1000) :|: TRUE f6(x7, x8) -> f7(x7, arith) :|: TRUE && arith = x8 - 1 f5(x9, x10) -> f6(x9, x10) :|: x10 > 1 f7(x11, x12) -> f5(x11, x12) :|: TRUE f5(x13, x14) -> f8(x13, x14) :|: x14 <= 1 f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f3(x17, x18) -> f4(x17, x18) :|: x17 <= 10 f9(x19, x20) -> f3(x19, x20) :|: TRUE f3(x21, x22) -> f10(x21, x22) :|: x21 > 10 Start term: f1(x1, x2) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x17, x18) -> f4(x17, x18) :|: x17 <= 10 f9(x19, x20) -> f3(x19, x20) :|: TRUE f8(x23, x24) -> f9(x25, x24) :|: TRUE && x25 = x23 + 1 f5(x13, x14) -> f8(x13, x14) :|: x14 <= 1 f7(x11, x12) -> f5(x11, x12) :|: TRUE f6(x7, x8) -> f7(x7, arith) :|: TRUE && arith = x8 - 1 f5(x9, x10) -> f6(x9, x10) :|: x10 > 1 f4(x5, x6) -> f5(x5, 1000) :|: TRUE ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x13:0, x14:0) -> f5(x13:0 + 1, 1000) :|: x14:0 < 2 && x13:0 < 10 f5(x9:0, x10:0) -> f5(x9:0, x10:0 - 1) :|: x10:0 > 1 ---------------------------------------- (7) RankingReductionPairProof (EQUIVALENT) Interpretation: [ f5 ] = -1*f5_1 The following rules are decreasing: f5(x13:0, x14:0) -> f5(x13:0 + 1, 1000) :|: x14:0 < 2 && x13:0 < 10 The following rules are bounded: f5(x13:0, x14:0) -> f5(x13:0 + 1, 1000) :|: x14:0 < 2 && x13:0 < 10 ---------------------------------------- (8) Obligation: Rules: f5(x9:0, x10:0) -> f5(x9:0, x10:0 - 1) :|: x10:0 > 1 ---------------------------------------- (9) IntTRSUnneededArgumentFilterProof (EQUIVALENT) Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: f5(x1, x2) -> f5(x2) ---------------------------------------- (10) Obligation: Rules: f5(x10:0) -> f5(x10:0 - 1) :|: x10:0 > 1 ---------------------------------------- (11) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f5(x)] = x The following rules are decreasing: f5(x10:0) -> f5(x10:0 - 1) :|: x10:0 > 1 The following rules are bounded: f5(x10:0) -> f5(x10:0 - 1) :|: x10:0 > 1 ---------------------------------------- (12) YES