/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1158 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(x, y) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f4(x4, x5) -> f5(x4, 1) :|: TRUE
f6(x6, x7) -> f7(x6, arith) :|: TRUE && arith = 2 * x7
f5(x8, x9) -> f6(x8, x9) :|: x8 > x9 && x9 > 0
f7(x10, x11) -> f5(x10, x11) :|: TRUE
f5(x12, x13) -> f8(x12, x13) :|: x12 <= x13
f5(x22, x23) -> f8(x22, x23) :|: x23 <= 0
f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 1
f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0 && x17 > 0
f9(x18, x19) -> f3(x18, x19) :|: TRUE
f3(x20, x21) -> f10(x20, x21) :|: x20 < 0
f3(x27, x28) -> f10(x27, x28) :|: x28 <= 0
Start term: f1(x, y)

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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)
   (f7_2,7)
   (f8_2,8)
   (f9_2,9)
   (f10_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;
assume(0 = 0);
x0 := oldX0;
x1 := 1;
TO: 5;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0 && oldX2 = 2 * oldX1);
x0 := oldX0;
x1 := oldX2;
TO: 7;

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

FROM: 7;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 5;

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

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

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

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

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

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

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


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(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 6, 9, 10, 11, 21, 22, 26, 27 using the following rank functions:
    - Rank function 1:
      RF for loc. 9: -1+3*x0
      RF for loc. 10: -2+3*x0
      RF for loc. 11: -1+3*x0
      RF for loc. 15: 3*x0
      Bound for (chained) transitions 10: -1
      Bound for (chained) transitions 11: -1
      Bound for (chained) transitions 21, 22: -2
      Bound for (chained) transitions 26: 0
      Bound for (chained) transitions 27: 0
    - Rank function 2:
      RF for loc. 9: oldX1+4*x0-4*x1
      RF for loc. 11: oldX1+4*x0-4*x1
      Bound for (chained) transitions 9: 5
    - Rank function 3:
      RF for loc. 9: 1
      RF for loc. 11: 0
      Bound for (chained) transitions 6: 1

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(6)
YES