/export/starexec/sandbox2/solver/bin/starexec_run_c /export/starexec/sandbox2/benchmark/theBenchmark.c /export/starexec/sandbox2/output/output_files


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.c
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given C Problem could be disproven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) IRS2T2 [EQUIVALENT, 0 ms]
(4) T2IntSys
(5) T2 [COMPLETE, 966 ms]
(6) NO


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(0)
Obligation:
c file /export/starexec/sandbox2/benchmark/theBenchmark.c
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(1) CToIRSProof (EQUIVALENT)
Parsed C Integer Program as IRS.
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(2)
Obligation:
Rules:
f1(i, b) -> f2(x_1, b) :|: TRUE
f3(x, x1) -> f4(arith, x1) :|: TRUE && arith = x - 2
f2(x2, x3) -> f3(x2, x3) :|: x2 < 1 && x2 < 0
f2(x8, x9) -> f3(x8, x9) :|: x8 < 1 && x8 > 0
f2(x10, x11) -> f3(x10, x11) :|: x10 > 1 && x10 < 0
f2(x12, x13) -> f3(x12, x13) :|: x12 > 1 && x12 > 0
f4(x4, x5) -> f2(x4, x5) :|: TRUE
f2(x6, x7) -> f5(x6, x7) :|: x6 = 1
f2(x14, x15) -> f5(x14, x15) :|: x14 = 0
Start term: f1(i, b)

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(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)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0);
x0 := oldX2;
x1 := oldX1;
TO: 2;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(2 - oldX0);
assume(0 = 0 && oldX2 = oldX0 - 2);
x0 := -(2 - oldX0);
x1 := oldX1;
TO: 4;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 1 && oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 1 && oldX0 > 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > 1 && oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > 1 && oldX0 > 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 = 1);
x0 := oldX0;
x1 := oldX1;
TO: 5;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 5;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 7: x0-oldX2 <= 0 and x0+1 <= 0 and x0 <= 0

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(6)
NO