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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1348 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(x_1, j) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f5(x3, x4) -> f8(arith, x4) :|: TRUE && arith = x3 + 4
f6(x21, x22) -> f9(x21, x23) :|: TRUE && x23 = x22 + 1
f9(x24, x25) -> f10(x26, x25) :|: TRUE && x26 = x24 + 2
f4(x9, x10) -> f5(x9, x10) :|: x9 < x10
f4(x11, x12) -> f6(x11, x12) :|: x11 >= x12
f8(x13, x14) -> f7(x13, x14) :|: TRUE
f10(x15, x16) -> f7(x15, x16) :|: TRUE
f3(x17, x18) -> f4(x17, x18) :|: TRUE
f7(x19, x20) -> f3(x19, x20) :|: TRUE
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)
   (f6_2,6)
   (f9_2,7)
   (f10_2,8)
   (f4_2,9)
   (f7_2,10)

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

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX1 + 1));
assume(0 = 0 && oldX2 = oldX1 + 1);
x0 := oldX0;
x1 := -(-(oldX1 + 1));
TO: 7;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX0 + 2));
assume(0 = 0 && oldX2 = oldX0 + 2);
x0 := -(-(oldX0 + 2));
x1 := oldX1;
TO: 8;

FROM: 9;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 4;

FROM: 9;
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: 10;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 10;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;

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


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 7: oldX2-x1 <= 0 and 0 <= 0

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(6)
NO