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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1328 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(bob, samantha, temp) -> f2(13, samantha, temp) :|: TRUE
f2(x, x1, x2) -> f3(x, 17, x2) :|: TRUE
f4(x3, x4, x5) -> f5(x3, x4, x3) :|: TRUE
f5(x6, x7, x8) -> f6(x7, x7, x8) :|: TRUE
f6(x9, x10, x11) -> f7(x9, x11, x11) :|: TRUE
f3(x12, x13, x14) -> f4(x12, x13, x14) :|: x12 + x13 < 100
f7(x15, x16, x17) -> f3(x15, x16, x17) :|: TRUE
f3(x18, x19, x20) -> f8(x18, x19, x20) :|: x18 + x19 >= 100
Start term: f1(bob, samantha, temp)

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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;
assume(0 = 0);
x0 := 13;
x1 := oldX1;
x2 := oldX2;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := 17;
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;
assume(0 = 0);
x0 := oldX1;
x1 := oldX1;
x2 := oldX2;
TO: 6;

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

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 + oldX1 < 100);
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 + oldX1 >= 100);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 8;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: oldX0 == 13 and x0 == 13 and x1 == 17 and x2 == 17 or oldX0 == 17 and x0 == 17 and x1 == 13 and x2 == 13

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(6)
NO