/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1158 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:
f6(x) -> f9(arith) :|: TRUE && arith = 0 - 5
f10(x1) -> f13(35) :|: TRUE
f11(x19) -> f14(x20) :|: TRUE && x20 = x19 - 1
f7(x3) -> f10(x3) :|: x3 > 30
f7(x4) -> f11(x4) :|: x4 <= 30
f13(x5) -> f12(x5) :|: TRUE
f14(x6) -> f12(x6) :|: TRUE
f3(x7) -> f6(x7) :|: x7 < 0
f3(x8) -> f7(x8) :|: x8 >= 0
f9(x9) -> f8(x9) :|: TRUE
f12(x10) -> f8(x10) :|: TRUE
f4(x11) -> f15(0) :|: TRUE
f2(x12) -> f3(x12) :|: 0 - 5 <= x12 && x12 <= 35
f2(x13) -> f4(x13) :|: 0 - 5 > x13
f2(x21) -> f4(x21) :|: x21 > 35
f8(x14) -> f5(x14) :|: TRUE
f15(x15) -> f5(x15) :|: TRUE
f1(x16) -> f2(x16) :|: x16 < 0
f1(x22) -> f2(x22) :|: x22 > 0
f5(x17) -> f1(x17) :|: TRUE
f1(x18) -> f16(x18) :|: x18 = 0
Start term: f1(x)

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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:

   (f6_1,1)
   (f9_1,2)
   (f10_1,3)
   (f13_1,4)
   (f11_1,5)
   (f14_1,6)
   (f7_1,7)
   (f12_1,8)
   (f3_1,9)
   (f8_1,10)
   (f4_1,11)
   (f15_1,12)
   (f2_1,13)
   (f5_1,14)
   (f1_1,15)
   (f16_1,16)

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(4)
Obligation:
START: 15;

FROM: 1;
oldX0 := x0;
oldX1 := -(5);
assume(0 = 0 && oldX1 = 0 - 5);
x0 := -(5);
TO: 2;

FROM: 3;
oldX0 := x0;
assume(0 = 0);
x0 := 35;
TO: 4;

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

FROM: 7;
oldX0 := x0;
assume(oldX0 > 30);
x0 := oldX0;
TO: 3;

FROM: 7;
oldX0 := x0;
assume(oldX0 <= 30);
x0 := oldX0;
TO: 5;

FROM: 4;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 8;

FROM: 6;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 8;

FROM: 9;
oldX0 := x0;
assume(oldX0 < 0);
x0 := oldX0;
TO: 1;

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

FROM: 2;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 10;

FROM: 8;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 10;

FROM: 11;
oldX0 := x0;
assume(0 = 0);
x0 := 0;
TO: 12;

FROM: 13;
oldX0 := x0;
assume(0 - 5 <= oldX0 && oldX0 <= 35);
x0 := oldX0;
TO: 9;

FROM: 13;
oldX0 := x0;
assume(0 - 5 > oldX0);
x0 := oldX0;
TO: 11;

FROM: 13;
oldX0 := x0;
assume(oldX0 > 35);
x0 := oldX0;
TO: 11;

FROM: 10;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 14;

FROM: 12;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 14;

FROM: 15;
oldX0 := x0;
assume(oldX0 < 0);
x0 := oldX0;
TO: 13;

FROM: 15;
oldX0 := x0;
assume(oldX0 > 0);
x0 := oldX0;
TO: 13;

FROM: 14;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 15;

FROM: 15;
oldX0 := x0;
assume(oldX0 = 0);
x0 := oldX0;
TO: 16;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 20: oldX0 == 35 and x0 == 35

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(6)
NO