/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1425 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(x, y, oldx) -> f2(x_1, y, oldx) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x10 - x9
f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 + x24
f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 < 5
f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE
f3(x20, x21, x22) -> f8(x20, x21, x22) :|: x20 >= 5
Start term: f1(x, y, oldx)

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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)

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

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := -(oldX1 - oldX2);
assume(0 = 0 && oldX3 = oldX2 - oldX1);
x0 := -(oldX1 - oldX2);
x1 := oldX1;
x2 := oldX2;
TO: 6;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := -(-(oldX2 + oldX1));
assume(0 = 0 && oldX3 = oldX2 + oldX1);
x0 := oldX0;
x1 := -(-(oldX2 + oldX1));
x2 := oldX2;
TO: 7;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 < 5);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 4;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 >= 5);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 8;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: oldX0 == 0 and oldX3 == 0 and x0 == 0 and x1 == 0 and x2 == 0

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(6)
NO