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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1594 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
f5(x2, x3) -> f8(arith, x3) :|: TRUE && arith = x2 + 1
f8(x24, x25) -> f9(x24, x26) :|: TRUE && x26 = x25 - 1
f6(x27, x28) -> f10(x29, x28) :|: TRUE && x29 = x27 - 1
f10(x30, x31) -> f11(x30, x32) :|: TRUE && x32 = x31 + 1
f4(x10, x11) -> f5(x10, x11) :|: 51 < x11
f4(x12, x13) -> f6(x12, x13) :|: 51 >= x13
f9(x14, x15) -> f7(x14, x15) :|: TRUE
f11(x16, x17) -> f7(x16, x17) :|: TRUE
f3(x18, x19) -> f4(x18, x19) :|: x18 < x19
f7(x20, x21) -> f3(x20, x21) :|: TRUE
f3(x22, x23) -> f12(x22, x23) :|: x22 >= x23
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)
   (f5_2,4)
   (f8_2,5)
   (f9_2,6)
   (f6_2,7)
   (f10_2,8)
   (f11_2,9)
   (f4_2,10)
   (f7_2,11)
   (f12_2,12)

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

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

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(1 - oldX0);
assume(0 = 0 && oldX2 = oldX0 - 1);
x0 := -(1 - oldX0);
x1 := oldX1;
TO: 8;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX1 + 1));
assume(0 = 0 && oldX2 = oldX1 + 1);
x0 := oldX0;
x1 := -(-(oldX1 + 1));
TO: 9;

FROM: 10;
oldX0 := x0;
oldX1 := x1;
assume(51 < oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 4;

FROM: 10;
oldX0 := x0;
oldX1 := x1;
assume(51 >= oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 7;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 11;

FROM: 9;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 11;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 10;

FROM: 11;
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: 12;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 8: 1 <= x0 and x0 <= 1 and 99 <= x1 and x1 <= 99 and x0-x1 <= -98 and x1-x0 <= 98 and x0+x1 <= 100 and 100 <= x0+x1 and -x1+x0+1 <= 0

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(6)
NO