NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.c
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1347 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(x, y) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 + x5
f3(x6, x7) -> f4(x6, x7) :|: x6 > 0
f5(x8, x9) -> f3(x8, x9) :|: TRUE
f3(x10, x11) -> f6(x10, x11) :|: x10 <= 0
Start term: f1(x, y)

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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)
   (f6_2,6)

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

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

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

FROM: 5;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 <= 0);
x0 := oldX0;
x1 := oldX1;
TO: 6;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: oldX2 == 0 and x0 == 1 and x1 == 0

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(6)
NO