NO
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1243 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) -> f2(x_1) :|: TRUE
f3(x1) -> f4(arith) :|: TRUE && arith = x1 + x2
f2(x3) -> f3(x3) :|: x3 >= 0
f4(x4) -> f2(x4) :|: TRUE
f2(x5) -> f5(x5) :|: x5 < 0
Start term: f1(x)

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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_1,1)
   (f2_1,2)
   (f3_1,3)
   (f4_1,4)
   (f5_1,5)

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(4)
Obligation:
START: 1;

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

FROM: 3;
oldX0 := x0;
oldX1 := nondet();
oldX2 := nondet();
assume(0 = 0 && oldX1 = oldX0 + oldX2);
x0 := oldX1;
TO: 4;

FROM: 2;
oldX0 := x0;
assume(oldX0 >= 0);
x0 := oldX0;
TO: 3;

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

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


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: oldX1 == 0 and oldX2 == 0 and x0 == 0

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(6)
NO