/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) IRS2T2 [EQUIVALENT, 0 ms] (4) T2IntSys (5) T2 [EQUIVALENT, 1386 ms] (6) 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(arith, x5) :|: TRUE && arith = x4 - 1 f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = 0 - 2 * x15 f3(x8, x9) -> f4(x8, x9) :|: x8 + x9 > 0 f6(x10, x11) -> f3(x10, x11) :|: TRUE f3(x12, x13) -> f7(x12, x13) :|: x12 + x13 <= 0 Start term: f1(x, y) ---------------------------------------- (3) IRS2T2 (EQUIVALENT) Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs: (f1_2,1) (f2_2,2) (f3_2,3) (f4_2,4) (f5_2,5) (f6_2,6) (f7_2,7) ---------------------------------------- (4) Obligation: START: 1; FROM: 1; oldX0 := x0; oldX1 := x1; oldX2 := nondet(); assume(0 = 0); x0 := oldX2; x1 := oldX1; TO: 2; FROM: 2; oldX0 := x0; oldX1 := x1; oldX2 := nondet(); assume(0 = 0); x0 := oldX0; x1 := oldX2; TO: 3; FROM: 4; oldX0 := x0; oldX1 := x1; oldX2 := -(1 - oldX0); assume(0 = 0 && oldX2 = oldX0 - 1); x0 := -(1 - oldX0); x1 := oldX1; TO: 5; FROM: 5; oldX0 := x0; oldX1 := x1; oldX2 := nondet(); assume(0 = 0 && oldX2 = 0 - 2 * oldX1); x0 := oldX0; x1 := oldX2; TO: 6; FROM: 3; oldX0 := x0; oldX1 := x1; assume(oldX0 + oldX1 > 0); x0 := oldX0; x1 := oldX1; TO: 4; FROM: 6; oldX0 := x0; oldX1 := x1; assume(0 = 0); x0 := oldX0; x1 := oldX1; TO: 3; FROM: 3; oldX0 := x0; oldX1 := x1; assume(oldX0 + oldX1 <= 0); x0 := oldX0; x1 := oldX1; TO: 7; ---------------------------------------- (5) T2 (EQUIVALENT) No proof given by T2 ---------------------------------------- (6) YES