/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 52 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/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, res) -> f2(x_1, y, res) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, 0) :|: TRUE
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x8 - x9
f6(x23, x24, x25) -> f7(x23, x24, x26) :|: TRUE && x26 = x25 + 1
f4(x14, x15, x16) -> f5(x14, x15, x16) :|: x14 >= x15 && x15 > 0
f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE
f4(x20, x21, x22) -> f8(x20, x21, x22) :|: x20 < x21
f4(x27, x28, x29) -> f8(x27, x28, x29) :|: x28 <= 0
Start term: f1(x, y, res)

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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(x14, x15, x16) -> f5(x14, x15, x16) :|: x14 >= x15 && x15 > 0
f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE
f6(x23, x24, x25) -> f7(x23, x24, x26) :|: TRUE && x26 = x25 + 1
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x8 - x9

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x23:0, x24:0, x25:0) -> f6(x23:0 - x24:0, x24:0, x25:0 + 1) :|: x24:0 <= x23:0 && x24:0 > 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:

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

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(8)
Obligation:
Rules:
f6(x23:0, x24:0) -> f6(x23:0 - x24:0, x24:0) :|: x24:0 <= x23:0 && x24:0 > 0

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

The following rules are decreasing:
f6(x23:0, x24:0) -> f6(x23:0 - x24:0, x24:0) :|: x24:0 <= x23:0 && x24:0 > 0
The following rules are bounded:
f6(x23:0, x24:0) -> f6(x23:0 - x24:0, x24:0) :|: x24:0 <= x23:0 && x24:0 > 0

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(10)
YES