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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1533 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
f3(x1, x2) -> f4(x1, 1) :|: TRUE
f5(x3, x4) -> f6(x3, arith) :|: TRUE && arith = 2 * x4
f4(x5, x6) -> f5(x5, x6) :|: x5 > x6
f6(x7, x8) -> f4(x7, x8) :|: TRUE
f4(x9, x10) -> f7(x9, x10) :|: x9 <= x10
f7(x19, x20) -> f8(x21, x20) :|: TRUE && x21 = x19 - 1
f2(x13, x14) -> f3(x13, x14) :|: x13 >= 0
f8(x15, x16) -> f2(x15, x16) :|: TRUE
f2(x17, x18) -> f9(x17, x18) :|: x17 < 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)

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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;
assume(0 = 0);
x0 := oldX0;
x1 := 1;
TO: 4;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := nondet();
assume(0 = 0 && oldX2 = 2 * oldX1);
x0 := oldX0;
x1 := oldX2;
TO: 6;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 5;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 4;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 <= oldX1);
x0 := oldX0;
x1 := oldX1;
TO: 7;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(1 - oldX0);
assume(0 = 0 && oldX2 = oldX0 - 1);
x0 := -(1 - oldX0);
x1 := oldX1;
TO: 8;

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

FROM: 8;
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: 9;


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(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 9, 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: 1+2*x0
      Bound for (chained) transitions 17: 1
      Bound for (chained) transitions 18: 1
    - Rank function 2:
      RF for loc. 8: 0
      RF for loc. 9: 0
      RF for loc. 13: -1
      Bound for (chained) transitions 9, 14: 0
Used the following cutpoint-specific lexicographic rank functions:
 * For cutpoint 8, used the following rank functions/bounds (in descending priority order):
    - RF -x1+oldX0, bound 1

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(6)
YES