/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 55 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 19 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) RankingReductionPairProof [EQUIVALENT, 0 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, y, oldx) -> f2(x_1, y, oldx) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1
f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1
f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0 && x15 >= 0
f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE
f3(x20, x21, x22) -> f8(x20, x21, x22) :|: x20 < 0
f3(x27, x28, x29) -> f8(x27, x28, x29) :|: x28 < 0
Start term: f1(x, y, 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:
f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0 && x15 >= 0
f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE
f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1
f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x8:0, x9:0, x10:0) -> f5(x9:0 - 1, x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 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, x3) -> f5(x2, x3)

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(8)
Obligation:
Rules:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0

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(9) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = 1/2*f5_1 + 1/2*f5_2

The following rules are decreasing:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0

The following rules are bounded:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x9:0 > 0 && x10:0 > 0


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(10)
YES