/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1043 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(i) -> f2(x_1) :|: TRUE
f3(x) -> f4(arith) :|: TRUE && arith = x - 1
f2(x1) -> f3(x1) :|: x1 > 5
f4(x2) -> f2(x2) :|: TRUE
f2(x3) -> f5(x3) :|: x3 <= 5
Start term: f1(i)

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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_1,1)
   (f2_1,2)
   (f3_1,3)
   (f4_1,4)
   (f5_1,5)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := nondet();
assume(0 = 0);
x0 := oldX1;
TO: 2;

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

FROM: 2;
oldX0 := x0;
assume(oldX0 > 5);
x0 := oldX0;
TO: 3;

FROM: 4;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 2;

FROM: 2;
oldX0 := x0;
assume(oldX0 <= 5);
x0 := oldX0;
TO: 5;


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(5) T2 (EQUIVALENT)
Initially, performed program simplifications using lexicographic rank functions:
 * Removed transitions 4, 7, 8 using the following rank functions:
    - Rank function 1:
      RF for loc. 6: 1+2*x0
      RF for loc. 7: 2*x0
      Bound for (chained) transitions 7: 12
      Bound for (chained) transitions 8: 12
    - Rank function 2:
      RF for loc. 6: 1
      RF for loc. 7: 0
      Bound for (chained) transitions 4: 1

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(6)
YES