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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1524 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(i) -> f2(3) :|: TRUE
f4(x) -> f7(arith) :|: TRUE && arith = x + 3
f8(x14) -> f11(x15) :|: TRUE && x15 = x14 - 2
f9(x16) -> f12(x17) :|: TRUE && x17 = x16 + 1
f5(x3) -> f8(x3) :|: x3 > 10
f5(x4) -> f9(x4) :|: x4 <= 10
f11(x5) -> f10(x5) :|: TRUE
f12(x6) -> f10(x6) :|: TRUE
f3(x7) -> f4(x7) :|: x7 > 5
f3(x8) -> f5(x8) :|: x8 <= 5
f7(x9) -> f6(x9) :|: TRUE
f10(x10) -> f6(x10) :|: TRUE
f2(x11) -> f3(x11) :|: x11 >= 3
f6(x12) -> f2(x12) :|: TRUE
f2(x13) -> f13(x13) :|: x13 < 3
Start term: f1(i)

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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)
   (f4_1,3)
   (f7_1,4)
   (f8_1,5)
   (f11_1,6)
   (f9_1,7)
   (f12_1,8)
   (f5_1,9)
   (f10_1,10)
   (f3_1,11)
   (f6_1,12)
   (f13_1,13)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
assume(0 = 0);
x0 := 3;
TO: 2;

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

FROM: 5;
oldX0 := x0;
oldX1 := -(2 - oldX0);
assume(0 = 0 && oldX1 = oldX0 - 2);
x0 := -(2 - oldX0);
TO: 6;

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

FROM: 9;
oldX0 := x0;
assume(oldX0 > 10);
x0 := oldX0;
TO: 5;

FROM: 9;
oldX0 := x0;
assume(oldX0 <= 10);
x0 := oldX0;
TO: 7;

FROM: 6;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 10;

FROM: 8;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 10;

FROM: 11;
oldX0 := x0;
assume(oldX0 > 5);
x0 := oldX0;
TO: 3;

FROM: 11;
oldX0 := x0;
assume(oldX0 <= 5);
x0 := oldX0;
TO: 9;

FROM: 4;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 12;

FROM: 10;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 12;

FROM: 2;
oldX0 := x0;
assume(oldX0 >= 3);
x0 := oldX0;
TO: 11;

FROM: 12;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 2;

FROM: 2;
oldX0 := x0;
assume(oldX0 < 3);
x0 := oldX0;
TO: 13;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 10: 6 <= oldX0 and 6 <= x0 and x0-oldX0 <= 0 and oldX0-x0 <= 0 and 12 <= x0+oldX0 and 0 <= 0

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(6)
NO