/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1373 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(q, y) -> f2(x_1, y) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f4(x3, x4) -> f5(arith, x4) :|: TRUE && arith = x3 - x4
f5(x13, x14) -> f6(x13, x15) :|: TRUE && x15 = x14 + 1
f3(x7, x8) -> f4(x7, x8) :|: x7 > 0
f6(x9, x10) -> f3(x9, x10) :|: TRUE
f3(x11, x12) -> f7(x11, x12) :|: x11 <= 0
Start term: f1(q, 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)

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

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

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := -(-(oldX1 + 1));
assume(0 = 0 && oldX2 = oldX1 + 1);
x0 := oldX0;
x1 := -(-(oldX1 + 1));
TO: 6;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 > 0);
x0 := oldX0;
x1 := oldX1;
TO: 4;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 <= 0);
x0 := oldX0;
x1 := oldX1;
TO: 7;


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(5) T2 (EQUIVALENT)
No proof given by T2
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(6)
YES