YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.c
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1284 ms]
(6) YES


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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(x, y) -> f2(x_1, y) :|: TRUE
f3(x1, x2) -> f4(arith, x2) :|: TRUE && arith = x1 + 1
f4(x3, x4) -> f5(x3, 1) :|: TRUE
f6(x21, x22) -> f7(x21, x23) :|: TRUE && x23 = x22 + 1
f5(x7, x8) -> f6(x7, x8) :|: x7 >= x8
f7(x9, x10) -> f5(x9, x10) :|: TRUE
f5(x11, x12) -> f8(x11, x12) :|: x11 < x12
f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 2
f2(x15, x16) -> f3(x15, x16) :|: x15 >= 0
f9(x17, x18) -> f2(x17, x18) :|: TRUE
f2(x19, x20) -> f10(x19, x20) :|: x19 < 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: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX0 + 1));
assume(0 = 0 && oldX2 = oldX0 + 1);
x0 := -(-(oldX0 + 1));
x1 := oldX1;
TO: 4;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := 1;
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: 5;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 >= oldX1);
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: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(2 - oldX0);
assume(0 = 0 && oldX2 = oldX0 - 2);
x0 := -(2 - oldX0);
x1 := oldX1;
TO: 9;

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

FROM: 9;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 10;


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(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 7, 9, 10, 11, 14, 17, 18 using the following rank functions:
    - Rank function 1:
      RF for loc. 8: 2*x0
      RF for loc. 9: 2*x0
      RF for loc. 13: 3+2*x0
      Bound for (chained) transitions 17: 3
      Bound for (chained) transitions 18: 3
    - Rank function 2:
      RF for loc. 8: x0-x1
      RF for loc. 9: x0-x1
      RF for loc. 13: 2+oldX0-x1
      Bound for (chained) transitions 10: 0
      Bound for (chained) transitions 11: 0
    - Rank function 3:
      RF for loc. 8: 0
      RF for loc. 9: -1
      RF for loc. 13: -1
      Bound for (chained) transitions 9, 14: 0
    - Rank function 4:
      RF for loc. 8: 0
      RF for loc. 9: -1
      Bound for (chained) transitions 7: 0

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(6)
YES