YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1113 ms]
(6) YES


----------------------------------------

(0)
Obligation:
c file /export/starexec/sandbox/benchmark/theBenchmark.c
----------------------------------------

(1) CToIRSProof (EQUIVALENT)
Parsed C Integer Program as IRS.
----------------------------------------

(2)
Obligation:
Rules:
f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1
f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1
f7(x3, x4) -> f8(x3, x5) :|: TRUE
f3(x6, x7) -> f4(x6, x7) :|: x7 < 0
f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0
f8(x10, x11) -> f6(x10, x11) :|: TRUE
f5(x12, x13) -> f6(x12, x13) :|: TRUE
f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0
f6(x16, x17) -> f1(x16, x17) :|: TRUE
f1(x18, x19) -> f9(x18, x19) :|: x18 < 0
f1(x23, x24) -> f9(x23, x24) :|: x24 < 0
Start term: f1(x, y)

----------------------------------------

(3) IRS2T2 (EQUIVALENT)
Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs:

   (f2_2,1)
   (f3_2,2)
   (f4_2,3)
   (f7_2,4)
   (f8_2,5)
   (f5_2,6)
   (f6_2,7)
   (f1_2,8)
   (f9_2,9)

----------------------------------------

(4)
Obligation:
START: 8;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(1 - oldX1);
assume(0 = 0 && oldX2 = oldX1 - 1);
x0 := oldX0;
x1 := -(1 - oldX1);
TO: 2;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(1 - oldX0);
assume(0 = 0 && oldX2 = oldX0 - 1);
x0 := -(1 - oldX0);
x1 := oldX1;
TO: 4;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX2;
TO: 5;

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

FROM: 2;
oldX0 := x0;
oldX1 := x1;
assume(oldX1 >= 0);
x0 := oldX0;
x1 := oldX1;
TO: 6;

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

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

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

FROM: 7;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 8;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
assume(oldX1 < 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;


----------------------------------------

(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 8, 9, 10, 14, 15, 18 using the following rank functions:
    - Rank function 1:
      RF for loc. 8: x0
      RF for loc. 9: x0
      RF for loc. 11: x0
      Bound for (chained) transitions 8: 0
    - Rank function 2:
      RF for loc. 8: 2+3*x1
      RF for loc. 9: 1+3*x1
      RF for loc. 11: 3*x1
      Bound for (chained) transitions 9: 2
      Bound for (chained) transitions 14: 0
    - Rank function 3:
      RF for loc. 8: -2
      RF for loc. 9: 0
      RF for loc. 11: -1
      Bound for (chained) transitions 10, 18: 0
      Bound for (chained) transitions 15: -1

----------------------------------------

(6)
YES