/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 42 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 34 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) TerminationGraphProcessor [EQUIVALENT, 0 ms]
(10) IntTRS
(11) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(12) IntTRS
(13) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(14) 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, z) -> f2(x_1, y, z) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE
f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10
f6(x12, x13, x14) -> f7(x12, x14, x14) :|: TRUE
f7(x27, x28, x29) -> f8(x27, x28, x30) :|: TRUE && x30 = 0 - x29 - 1
f4(x18, x19, x20) -> f5(x18, x19, x20) :|: x18 > 0
f8(x21, x22, x23) -> f4(x21, x22, x23) :|: TRUE
f4(x24, x25, x26) -> f9(x24, x25, x26) :|: x24 <= 0
Start term: f1(x, y, z)

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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 > 0
f8(x21, x22, x23) -> f4(x21, x22, x23) :|: TRUE
f7(x27, x28, x29) -> f8(x27, x28, x30) :|: TRUE && x30 = 0 - x29 - 1
f6(x12, x13, x14) -> f7(x12, x14, x14) :|: TRUE
f5(x9, x10, x11) -> f6(arith, x10, x11) :|: TRUE && arith = x9 + x10

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x12:0, x13:0, x14:0) -> f6(x12:0 + x14:0, x14:0, 0 - x14:0 - 1) :|: x12:0 > 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, x3)

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(8)
Obligation:
Rules:
f6(x12:0, x14:0) -> f6(x12:0 + x14:0, 0 - x14:0 - 1) :|: x12:0 > 0

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(9) TerminationGraphProcessor (EQUIVALENT)
Constructed the termination graph and obtained one non-trivial SCC.

f6(x12:0, x14:0) -> f6(x12:0 + x14:0, 0 - x14:0 - 1) :|: x12:0 > 0
has been transformed into
f6(x12:0, x14:0) -> f6(x12:0 + x14:0, 0 - x14:0 - 1) :|: x12:0 > 0 && x4 > 0.


f6(x12:0, x14:0) -> f6(x12:0 + x14:0, 0 - x14:0 - 1) :|: x12:0 > 0 && x4 > 0 and 
f6(x12:0, x14:0) -> f6(x12:0 + x14:0, 0 - x14:0 - 1) :|: x12:0 > 0 && x4 > 0
have been merged into the new rule
f6(x12, x13) -> f6(x12 + x13 + (0 - x13 - 1), 0 - (0 - x13 - 1) - 1) :|: x12 > 0 && x14 > 0 && (x12 + x13 > 0 && x15 > 0)


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(10)
Obligation:
Rules:
f6(x16, x17) -> f6(x16 + -1, x17) :|: TRUE && x16 >= 1 && x18 >= 1 && x16 + x17 >= 1 && x19 >= 1

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(11) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(12)
Obligation:
Rules:
f6(x16:0, x17:0) -> f6(x16:0 - 1, x17:0) :|: x16:0 + x17:0 >= 1 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0

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

The following rules are decreasing:
f6(x16:0, x17:0) -> f6(x16:0 - 1, x17:0) :|: x16:0 + x17:0 >= 1 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0
The following rules are bounded:
f6(x16:0, x17:0) -> f6(x16:0 - 1, x17:0) :|: x16:0 + x17:0 >= 1 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0

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(14)
YES