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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 46 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 42 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) RankingReductionPairProof [EQUIVALENT, 20 ms]
(10) YES


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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(x, oldx) -> f2(x_1, oldx) :|: TRUE
f3(x1, x2) -> f4(x1, x1) :|: TRUE
f4(x3, x4) -> f5(x5, x4) :|: TRUE
f2(x6, x7) -> f3(x6, x7) :|: x6 > 0 && x6 < 100 && x6 >= 2 * x7 + 10
f5(x8, x9) -> f2(x8, x9) :|: TRUE
f2(x10, x11) -> f6(x10, x11) :|: x10 < 2 * x11 + 10
f2(x12, x13) -> f6(x12, x13) :|: x12 <= 0
f2(x14, x15) -> f6(x14, x15) :|: x14 >= 100
Start term: f1(x, oldx)

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

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(4)
Obligation:
Rules:
f2(x6, x7) -> f3(x6, x7) :|: x6 > 0 && x6 < 100 && x6 >= 2 * x7 + 10
f5(x8, x9) -> f2(x8, x9) :|: TRUE
f4(x3, x4) -> f5(x5, x4) :|: TRUE
f3(x1, x2) -> f4(x1, x1) :|: TRUE

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f4(x3:0, x4:0) -> f4(x5:0, x5:0) :|: x5:0 > 0 && x5:0 < 100 && x5:0 >= 2 * x4:0 + 10

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

   f4(x1, x2) -> f4(x2)

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(8)
Obligation:
Rules:
f4(x4:0) -> f4(x5:0) :|: x5:0 > 0 && x5:0 < 100 && x5:0 >= 2 * x4:0 + 10

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(9) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f4 ] = -2/11*f4_1

The following rules are decreasing:
f4(x4:0) -> f4(x5:0) :|: x5:0 > 0 && x5:0 < 100 && x5:0 >= 2 * x4:0 + 10

The following rules are bounded:
f4(x4:0) -> f4(x5:0) :|: x5:0 > 0 && x5:0 < 100 && x5:0 >= 2 * x4:0 + 10


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(10)
YES