5.74/2.34 YES 5.74/2.35 proof of /export/starexec/sandbox/benchmark/theBenchmark.c 5.74/2.35 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 5.74/2.35 5.74/2.35 5.74/2.35 Termination of the given C Problem could be proven: 5.74/2.35 5.74/2.35 (0) C Problem 5.74/2.35 (1) CToIRSProof [EQUIVALENT, 0 ms] 5.74/2.35 (2) IntTRS 5.74/2.35 (3) TerminationGraphProcessor [SOUND, 21 ms] 5.74/2.35 (4) IntTRS 5.74/2.35 (5) IntTRSCompressionProof [EQUIVALENT, 34 ms] 5.74/2.35 (6) IntTRS 5.74/2.35 (7) TerminationGraphProcessor [EQUIVALENT, 0 ms] 5.74/2.35 (8) IntTRS 5.74/2.35 (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] 5.74/2.35 (10) IntTRS 5.74/2.35 (11) PolynomialOrderProcessor [EQUIVALENT, 7 ms] 5.74/2.35 (12) YES 5.74/2.35 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (0) 5.74/2.35 Obligation: 5.74/2.35 c file /export/starexec/sandbox/benchmark/theBenchmark.c 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (1) CToIRSProof (EQUIVALENT) 5.74/2.35 Parsed C Integer Program as IRS. 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (2) 5.74/2.35 Obligation: 5.74/2.35 Rules: 5.74/2.35 f1(x, y) -> f2(x_1, y) :|: TRUE 5.74/2.35 f2(x1, x2) -> f3(x1, x3) :|: TRUE 5.74/2.35 f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 + x5 5.74/2.35 f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = x15 - 1 5.74/2.35 f3(x8, x9) -> f4(x8, x9) :|: x8 >= 0 5.74/2.35 f6(x10, x11) -> f3(x10, x11) :|: TRUE 5.74/2.35 f3(x12, x13) -> f7(x12, x13) :|: x12 < 0 5.74/2.35 Start term: f1(x, y) 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (3) TerminationGraphProcessor (SOUND) 5.74/2.35 Constructed the termination graph and obtained one non-trivial SCC. 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (4) 5.74/2.35 Obligation: 5.74/2.35 Rules: 5.74/2.35 f3(x8, x9) -> f4(x8, x9) :|: x8 >= 0 5.74/2.35 f6(x10, x11) -> f3(x10, x11) :|: TRUE 5.74/2.35 f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = x15 - 1 5.74/2.35 f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 + x5 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (5) IntTRSCompressionProof (EQUIVALENT) 5.74/2.35 Compressed rules. 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (6) 5.74/2.35 Obligation: 5.74/2.35 Rules: 5.74/2.35 f5(x14:0, x15:0) -> f5(x14:0 + (x15:0 - 1), x15:0 - 1) :|: x14:0 > -1 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (7) TerminationGraphProcessor (EQUIVALENT) 5.74/2.35 Constructed the termination graph and obtained one non-trivial SCC. 5.74/2.35 5.74/2.35 f5(x14:0, x15:0) -> f5(x14:0 + (x15:0 - 1), x15:0 - 1) :|: x14:0 > -1 5.74/2.35 has been transformed into 5.74/2.35 f5(x14:0, x15:0) -> f5(x14:0 + (x15:0 - 1), x15:0 - 1) :|: x14:0 > -1 && x4 > -1. 5.74/2.35 5.74/2.35 5.74/2.35 f5(x14:0, x15:0) -> f5(x14:0 + (x15:0 - 1), x15:0 - 1) :|: x14:0 > -1 && x4 > -1 and 5.74/2.35 f5(x14:0, x15:0) -> f5(x14:0 + (x15:0 - 1), x15:0 - 1) :|: x14:0 > -1 && x4 > -1 5.74/2.35 have been merged into the new rule 5.74/2.35 f5(x12, x13) -> f5(x12 + (x13 - 1) + (x13 - 1 - 1), x13 - 1 - 1) :|: x12 > -1 && x14 > -1 && (x12 + (x13 - 1) > -1 && x15 > -1) 5.74/2.35 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (8) 5.74/2.35 Obligation: 5.74/2.35 Rules: 5.74/2.35 f5(x16, x17) -> f5(x16 + 2 * x17 + -3, x17 + -2) :|: TRUE && x16 >= 0 && x18 >= 0 && x16 + x17 >= 1 && x19 >= 0 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (9) IntTRSCompressionProof (EQUIVALENT) 5.74/2.35 Compressed rules. 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (10) 5.74/2.35 Obligation: 5.74/2.35 Rules: 5.74/2.35 f5(x16:0, x17:0) -> f5(x16:0 + 2 * x17:0 - 3, x17:0 - 2) :|: x16:0 + x17:0 >= 1 && x19:0 > -1 && x16:0 > -1 && x18:0 > -1 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (11) PolynomialOrderProcessor (EQUIVALENT) 5.74/2.35 Found the following polynomial interpretation: 5.74/2.35 [f5(x, x1)] = 2*x + x1^2 5.74/2.35 5.74/2.35 The following rules are decreasing: 5.74/2.35 f5(x16:0, x17:0) -> f5(x16:0 + 2 * x17:0 - 3, x17:0 - 2) :|: x16:0 + x17:0 >= 1 && x19:0 > -1 && x16:0 > -1 && x18:0 > -1 5.74/2.35 The following rules are bounded: 5.74/2.35 f5(x16:0, x17:0) -> f5(x16:0 + 2 * x17:0 - 3, x17:0 - 2) :|: x16:0 + x17:0 >= 1 && x19:0 > -1 && x16:0 > -1 && x18:0 > -1 5.74/2.35 5.74/2.35 ---------------------------------------- 5.74/2.35 5.74/2.35 (12) 5.74/2.35 YES 6.20/2.38 EOF