/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 49 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 31 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) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - x5
f5(x6, x7) -> f6(x6, x8) :|: TRUE
f3(x9, x10) -> f4(x9, x10) :|: x9 >= 0 && x10 >= 1
f6(x11, x12) -> f3(x11, x12) :|: TRUE
f3(x13, x14) -> f7(x13, x14) :|: x13 < 0
f3(x15, x16) -> f7(x15, x16) :|: x16 < 1
Start term: f1(x, y)

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

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(4)
Obligation:
Rules:
f3(x9, x10) -> f4(x9, x10) :|: x9 >= 0 && x10 >= 1
f6(x11, x12) -> f3(x11, x12) :|: TRUE
f5(x6, x7) -> f6(x6, x8) :|: TRUE
f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - x5

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x6:0, x7:0) -> f5(x6:0 - x8:0, x8:0) :|: x6:0 > -1 && x8:0 > 0

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

   f5(x1, x2) -> f5(x1)

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(8)
Obligation:
Rules:
f5(x6:0) -> f5(x6:0 - x8:0) :|: x6:0 > -1 && x8:0 > 0

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x)] = x

The following rules are decreasing:
f5(x6:0) -> f5(x6:0 - x8:0) :|: x6:0 > -1 && x8:0 > 0
The following rules are bounded:
f5(x6:0) -> f5(x6:0 - x8:0) :|: x6:0 > -1 && x8:0 > 0

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(10)
YES