YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.c
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given C Problem could be proven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) IRS2T2 [EQUIVALENT, 0 ms]
(4) T2IntSys
(5) T2 [EQUIVALENT, 1267 ms]
(6) YES


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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(y, z) -> f2(x_1, z) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f4(x3, x4) -> f5(arith, x4) :|: TRUE && arith = x3 - 1
f6(x5, x6) -> f9(x5, x7) :|: TRUE
f7(x24, x25) -> f10(x24, x26) :|: TRUE && x26 = x25 - 1
f5(x10, x11) -> f6(x10, x11) :|: x10 >= 0
f5(x12, x13) -> f7(x12, x13) :|: x12 < 0
f9(x14, x15) -> f8(x14, x15) :|: TRUE
f10(x16, x17) -> f8(x16, x17) :|: TRUE
f3(x18, x19) -> f4(x18, x19) :|: x19 >= 0
f8(x20, x21) -> f3(x20, x21) :|: TRUE
f3(x22, x23) -> f11(x22, x23) :|: x23 < 0
Start term: f1(y, z)

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

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

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

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

FROM: 5;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 >= 0);
x0 := oldX0;
x1 := oldX1;
TO: 6;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 8;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 10;

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

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX1 >= 0);
x0 := oldX0;
x1 := oldX1;
TO: 4;

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

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX1 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 11;


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(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 7, 8, 9, 12, 13, 16 using the following rank functions:
    - Rank function 1:
      RF for loc. 9: 2*x0
      RF for loc. 10: -1+2*x0
      RF for loc. 11: -1+2*x0
      Bound for (chained) transitions 7: 0
    - Rank function 2:
      RF for loc. 9: -1+3*x1
      RF for loc. 10: 1+3*x1
      RF for loc. 11: 3*x1
      Bound for (chained) transitions 8: -1
      Bound for (chained) transitions 12: 0
      Bound for (chained) transitions 13: 0
    - Rank function 3:
      RF for loc. 10: 1
      RF for loc. 11: 0
      Bound for (chained) transitions 9, 16: 1

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(6)
YES