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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1337 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) -> f2(i, 100) :|: TRUE
f2(x, x1) -> f3(0, x1) :|: TRUE
f4(x2, x3) -> f5(x2, arith) :|: TRUE && arith = x3 + 1
f5(x12, x13) -> f6(x14, x13) :|: TRUE && x14 = x12 + 1
f3(x6, x7) -> f4(x6, x7) :|: x6 < x7
f6(x8, x9) -> f3(x8, x9) :|: TRUE
f3(x10, x11) -> f7(x10, x11) :|: x10 >= x11
Start term: f1(i, j)

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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)
   (f7_2,7)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := 100;
TO: 2;

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

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX1 + 1));
assume(0 = 0 && oldX2 = oldX1 + 1);
x0 := oldX0;
x1 := -(-(oldX1 + 1));
TO: 5;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX0 + 1));
assume(0 = 0 && oldX2 = oldX0 + 1);
x0 := -(-(oldX0 + 1));
x1 := oldX1;
TO: 6;

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

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

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 >= oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 7;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: 0 <= x0 and 100 <= x1 and x0-x1 <= -100 and x1-x0 <= 100 and 100 <= x0+x1 and -x1+x0+1 <= 0

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(6)
NO