/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) TerminationGraphProcessor [SOUND, 53 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) 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(12, y, z) :|: TRUE
f2(x1, x2, x3) -> f3(x1, 0, x3) :|: TRUE
f3(x4, x5, x6) -> f4(x4, x5, x7) :|: TRUE
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x8 + 1
f6(x26, x27, x28) -> f7(x26, x29, x28) :|: TRUE && x29 = x27 - 1
f7(x30, x31, x32) -> f8(x30, x31, x33) :|: TRUE && x33 = x32 - 1
f4(x17, x18, x19) -> f5(x17, x18, x19) :|: x19 > 0
f8(x20, x21, x22) -> f4(x20, x21, x22) :|: TRUE
f4(x23, x24, x25) -> f9(x23, x24, x25) :|: x25 <= 0
Start term: f1(x, y, z)

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(3) TerminationGraphProcessor (SOUND)
Constructed the termination graph and obtained one non-trivial SCC.

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(4)
Obligation:
Rules:
f4(x17, x18, x19) -> f5(x17, x18, x19) :|: x19 > 0
f8(x20, x21, x22) -> f4(x20, x21, x22) :|: TRUE
f7(x30, x31, x32) -> f8(x30, x31, x33) :|: TRUE && x33 = x32 - 1
f6(x26, x27, x28) -> f7(x26, x29, x28) :|: TRUE && x29 = x27 - 1
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x8 + 1

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x26:0, x27:0, x28:0) -> f6(x26:0 + 1, x27:0 - 1, x28:0 - 1) :|: x28:0 > 1

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(7) IntTRSUnneededArgumentFilterProof (EQUIVALENT)
Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements:

   f6(x1, x2, x3) -> f6(x3)

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(8)
Obligation:
Rules:
f6(x28:0) -> f6(x28:0 - 1) :|: x28:0 > 1

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x)] = x

The following rules are decreasing:
f6(x28:0) -> f6(x28:0 - 1) :|: x28:0 > 1
The following rules are bounded:
f6(x28:0) -> f6(x28:0 - 1) :|: x28:0 > 1

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(10)
YES