YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1362 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) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f7(x4, x5) -> f8(x4, arith) :|: TRUE && arith = x5 + x4
f4(x6, x7) -> f7(x6, x7) :|: x6 > x7
f8(x8, x9) -> f4(x8, x9) :|: TRUE
f4(x10, x11) -> f9(x10, x11) :|: x10 <= x11
f3(x12, x13) -> f4(x12, x13) :|: x12 > 0
f3(x14, x15) -> f5(x14, x15) :|: x14 <= 0
f9(x16, x17) -> f6(x16, x17) :|: TRUE
f5(x18, x19) -> f6(x18, x19) :|: TRUE
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)
   (f7_2,4)
   (f8_2,5)
   (f4_2,6)
   (f9_2,7)
   (f5_2,8)
   (f6_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: 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 = oldX1 + oldX0);
x0 := oldX0;
x1 := -(-(oldX1 + oldX0));
TO: 5;

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

FROM: 5;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 6;

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

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

FROM: 3;
oldX0 := x0;
oldX1 := x1;
assume(oldX0 <= 0);
x0 := oldX0;
x1 := oldX1;
TO: 8;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
TO: 9;

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


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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 -x1+oldX0, bound 1

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(6)
YES