/export/starexec/sandbox2/solver/bin/starexec_run_c /export/starexec/sandbox2/benchmark/theBenchmark.c /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1183 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, z) -> f2(x_1, y, z) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE
f8(x9, x10, x11) -> f9(x9, x10, arith) :|: TRUE && arith = x11 + x10
f5(x12, x13, x14) -> f8(x12, x13, x14) :|: x12 >= x14
f9(x15, x16, x17) -> f5(x15, x16, x17) :|: TRUE
f5(x18, x19, x20) -> f10(x18, x19, x20) :|: x18 < x20
f4(x21, x22, x23) -> f5(x21, x22, x23) :|: x22 > 0
f4(x24, x25, x26) -> f6(x24, x25, x26) :|: x25 <= 0
f10(x27, x28, x29) -> f7(x27, x28, x29) :|: TRUE
f6(x30, x31, x32) -> f7(x30, x31, x32) :|: TRUE
Start term: f1(x, y, z)

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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_3,1)
   (f2_3,2)
   (f3_3,3)
   (f4_3,4)
   (f8_3,5)
   (f9_3,6)
   (f5_3,7)
   (f10_3,8)
   (f6_3,9)
   (f7_3,10)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX3;
x1 := oldX1;
x2 := oldX2;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX3;
x2 := oldX2;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX3;
TO: 4;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := -(-(oldX2 + oldX1));
assume(0 = 0 && oldX3 = oldX2 + oldX1);
x0 := oldX0;
x1 := oldX1;
x2 := -(-(oldX2 + oldX1));
TO: 6;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 >= oldX2);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 7;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 < oldX2);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 8;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX1 > 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 7;

FROM: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX1 <= 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 9;

FROM: 8;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 10;

FROM: 9;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 10;


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(5) T2 (EQUIVALENT)
Used the following cutpoint-specific lexicographic rank functions:
 * For cutpoint 7, used the following rank functions/bounds (in descending priority order):
    - RF -x2+x0, bound 0

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(6)
YES