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


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NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.c
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination of the given C Problem could be disproven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) IRS2T2 [EQUIVALENT, 0 ms]
(4) T2IntSys
(5) T2 [COMPLETE, 6563 ms]
(6) NO


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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, j, r) -> f2(x_1, j, r) :|: TRUE
f2(x, x1, x2) -> f3(x, x3, x2) :|: TRUE
f4(x4, x5, x6) -> f5(arith, x5, x6) :|: TRUE && arith = x4 - x7
f5(x24, x25, x26) -> f6(x24, x25, x27) :|: TRUE && x27 = x28 + 1
f6(x29, x30, x31) -> f7(x29, x32, x31) :|: TRUE && x32 = x30 + x31
f3(x15, x16, x17) -> f4(x15, x16, x17) :|: x15 - x16 >= 1
f7(x18, x19, x20) -> f3(x18, x19, x20) :|: TRUE
f3(x21, x22, x23) -> f8(x21, x22, x23) :|: x21 - x22 < 1
Start term: f1(i, j, r)

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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)
   (f5_3,5)
   (f6_3,6)
   (f7_3,7)
   (f8_3,8)

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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: 4;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
oldX4 := nondet();
assume(0 = 0 && oldX3 = oldX0 - oldX4);
x0 := oldX3;
x1 := oldX1;
x2 := oldX2;
TO: 5;

FROM: 5;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
oldX4 := nondet();
assume(0 = 0 && oldX3 = oldX4 + 1);
x0 := oldX0;
x1 := oldX1;
x2 := oldX3;
TO: 6;

FROM: 6;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := -(-(oldX1 + oldX2));
assume(0 = 0 && oldX3 = oldX1 + oldX2);
x0 := oldX0;
x1 := -(-(oldX1 + oldX2));
x2 := oldX2;
TO: 7;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 - oldX1 >= 1);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 4;

FROM: 7;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(0 = 0);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
assume(oldX0 - oldX1 < 1);
x0 := oldX0;
x1 := oldX1;
x2 := oldX2;
TO: 8;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 6: oldX2 == 0 and oldX3 == 0 and oldX4 == -1 and x0 == 1 and x1 == 0 and x2 == 0

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(6)
NO