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


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NO
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1267 ms]
(6) NO


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(0)
Obligation:
c file /export/starexec/sandbox/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, j, t) -> f2(x_1, j, t) :|: TRUE
f2(x, x1, x2) -> f3(x, x3, x2) :|: TRUE
f3(x4, x5, x6) -> f4(x4, x5, 0) :|: TRUE
f5(x7, x8, x9) -> f6(x7, x8, x7) :|: TRUE
f6(x10, x11, x12) -> f7(x11, x11, x12) :|: TRUE
f7(x13, x14, x15) -> f8(x13, x15, x15) :|: TRUE
f4(x16, x17, x18) -> f5(x16, x17, x18) :|: x16 < 0 && x17 < 0
f4(x25, x26, x27) -> f5(x25, x26, x27) :|: x25 < 0 && x26 > 0
f4(x28, x29, x30) -> f5(x28, x29, x30) :|: x28 > 0 && x29 < 0
f4(x31, x32, x33) -> f5(x31, x32, x33) :|: x31 > 0 && x32 > 0
f8(x19, x20, x21) -> f4(x19, x20, x21) :|: TRUE
f4(x22, x23, x24) -> f9(x22, x23, x24) :|: x22 = 0
f4(x34, x35, x36) -> f9(x34, x35, x36) :|: x35 = 0
Start term: f1(i, j, t)

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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_3,1)
   (f2_3,2)
   (f3_3,3)
   (f4_3,4)
   (f5_3,5)
   (f6_3,6)
   (f7_3,7)
   (f8_3,8)
   (f9_3,9)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX3;
x1 := oldX1;
x2 := oldX2;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX3;
x2 := oldX2;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := 0;
TO: 4;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX0;
TO: 6;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX1;
x1 := oldX1;
x2 := oldX2;
TO: 7;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX2;
x2 := oldX2;
TO: 8;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 < 0 && oldX1 < 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 < 0 && oldX1 > 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 > 0 && oldX1 < 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 > 0 && oldX1 > 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 4;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 9;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX1 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 9;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 7: oldX0 == 1 and oldX3 == 1 and x0 == 1 and x1 == 1 and x2 == 1

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(6)
NO