/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 38 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 36 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, tmp) -> f2(x_1, y, tmp) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE
f5(x9, x10, x11) -> f6(x9, x10, x9) :|: TRUE
f6(x12, x13, x14) -> f7(x13, x13, x14) :|: TRUE
f7(x15, x16, x17) -> f8(x15, x17, x17) :|: TRUE
f4(x18, x19, x20) -> f5(x18, x19, x20) :|: x18 > x19
f8(x21, x22, x23) -> f4(x21, x22, x23) :|: TRUE
f4(x24, x25, x26) -> f9(x24, x25, x26) :|: x24 <= x25
Start term: f1(x, y, tmp)

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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(x18, x19, x20) -> f5(x18, x19, x20) :|: x18 > x19
f8(x21, x22, x23) -> f4(x21, x22, x23) :|: TRUE
f7(x15, x16, x17) -> f8(x15, x17, x17) :|: TRUE
f6(x12, x13, x14) -> f7(x13, x13, x14) :|: TRUE
f5(x9, x10, x11) -> f6(x9, x10, x9) :|: TRUE

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x12:0, x13:0, x14:0) -> f6(x13:0, x14:0, x13:0) :|: x14:0 < x13:0

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

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

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(8)
Obligation:
Rules:
f6(x13:0, x14:0) -> f6(x14:0, x13:0) :|: x14:0 < x13:0

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x, x1)] = -1 + x - x1

The following rules are decreasing:
f6(x13:0, x14:0) -> f6(x14:0, x13:0) :|: x14:0 < x13:0
The following rules are bounded:
f6(x13:0, x14:0) -> f6(x14:0, x13:0) :|: x14:0 < x13:0

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(10)
YES