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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1377 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(y1, y2) -> f2(x_1, y2) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f8(x3, x4) -> f11(arith, x4) :|: TRUE && arith = x3 - x4
f9(x29, x30) -> f12(x29, x31) :|: TRUE && x31 = x30 - x29
f7(x7, x8) -> f8(x7, x8) :|: x7 > x8
f7(x9, x10) -> f9(x9, x10) :|: x9 <= x10
f11(x11, x12) -> f10(x11, x12) :|: TRUE
f12(x13, x14) -> f10(x13, x14) :|: TRUE
f4(x15, x16) -> f7(x15, x16) :|: x15 < x16
f4(x32, x33) -> f7(x32, x33) :|: x32 > x33
f10(x17, x18) -> f4(x17, x18) :|: TRUE
f4(x19, x20) -> f13(x19, x20) :|: x19 = x20
f3(x21, x22) -> f4(x21, x22) :|: x21 >= 0 && x22 >= 0
f3(x23, x24) -> f5(x23, x24) :|: x23 < 0
f3(x34, x35) -> f5(x34, x35) :|: x35 < 0
f13(x25, x26) -> f6(x25, x26) :|: TRUE
f5(x27, x28) -> f6(x27, x28) :|: TRUE
Start term: f1(y1, y2)

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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)
   (f8_2,4)
   (f11_2,5)
   (f9_2,6)
   (f12_2,7)
   (f7_2,8)
   (f10_2,9)
   (f4_2,10)
   (f13_2,11)
   (f5_2,12)
   (f6_2,13)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0);
x0 := oldX2;
x1 := oldX1;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX2;
TO: 3;

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

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

FROM: 8;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 4;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 <= oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 6;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;

FROM: 10;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 8;

FROM: 10;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 8;

FROM: 9;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 10;

FROM: 10;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 = oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 11;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 >= 0 && oldX1 >= 0);
x0 := oldX0;
x1 := oldX1;
TO: 10;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 12;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX1 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 12;

FROM: 11;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 13;

FROM: 12;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 13;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 12: oldX2 == 0 and x0 == 1 and x1 == 0

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(6)
NO