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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1487 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(a, b, am, bm) -> f2(a, b, a, bm) :|: TRUE
f2(x, x1, x2, x3) -> f3(x, x1, x2, x1) :|: TRUE
f5(x4, x5, x6, x7) -> f8(x4, x5, x6, arith) :|: TRUE && arith = x7 + x5
f6(x40, x41, x42, x43) -> f9(x40, x41, x44, x43) :|: TRUE && x44 = x42 + x40
f4(x12, x13, x14, x15) -> f5(x12, x13, x14, x15) :|: x14 > x15
f4(x16, x17, x18, x19) -> f6(x16, x17, x18, x19) :|: x18 <= x19
f8(x20, x21, x22, x23) -> f7(x20, x21, x22, x23) :|: TRUE
f9(x24, x25, x26, x27) -> f7(x24, x25, x26, x27) :|: TRUE
f3(x28, x29, x30, x31) -> f4(x28, x29, x30, x31) :|: x30 < x31
f3(x45, x46, x47, x48) -> f4(x45, x46, x47, x48) :|: x47 > x48
f7(x32, x33, x34, x35) -> f3(x32, x33, x34, x35) :|: TRUE
f3(x36, x37, x38, x39) -> f10(x36, x37, x38, x39) :|: x38 = x39
Start term: f1(a, b, am, bm)

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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_4,1)
   (f2_4,2)
   (f3_4,3)
   (f5_4,4)
   (f8_4,5)
   (f6_4,6)
   (f9_4,7)
   (f4_4,8)
   (f7_4,9)
   (f10_4,10)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX0;
x3 := oldX3;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX1;
TO: 3;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
oldX4 := -(-(oldX3 + oldX1));
assume(0 = 0 && oldX4 = oldX3 + oldX1);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := -(-(oldX3 + oldX1));
TO: 5;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
oldX4 := -(-(oldX2 + oldX0));
assume(0 = 0 && oldX4 = oldX2 + oldX0);
x0 := oldX0;
x1 := oldX1;
x2 := -(-(oldX2 + oldX0));
x3 := oldX3;
TO: 7;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(oldX2 > oldX3);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 4;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(oldX2 <= oldX3);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 6;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 9;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 9;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(oldX2 < oldX3);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 8;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(oldX2 > oldX3);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 8;

FROM: 9;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := x3;
assume(oldX2 = oldX3);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
x3 := oldX3;
TO: 10;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 8: oldX0 == 0 and oldX1 == 1 and x0 == 0 and x1 == 1 and x2 == 0 and x3 == 1

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(6)
NO