/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 46 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 26 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 12 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:
f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1
f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1
f7(x3, x4) -> f8(x3, x5) :|: TRUE
f3(x6, x7) -> f4(x6, x7) :|: x7 < 0
f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0
f8(x10, x11) -> f6(x10, x11) :|: TRUE
f5(x12, x13) -> f6(x12, x13) :|: TRUE
f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0
f6(x16, x17) -> f1(x16, x17) :|: TRUE
f1(x18, x19) -> f9(x18, x19) :|: x18 < 0
f1(x23, x24) -> f9(x23, x24) :|: x24 < 0
Start term: f1(x, y)

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

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(4)
Obligation:
Rules:
f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0
f6(x16, x17) -> f1(x16, x17) :|: TRUE
f8(x10, x11) -> f6(x10, x11) :|: TRUE
f7(x3, x4) -> f8(x3, x5) :|: TRUE
f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1
f3(x6, x7) -> f4(x6, x7) :|: x7 < 0
f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1
f5(x12, x13) -> f6(x12, x13) :|: TRUE
f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f3(x8:0, x9:0) -> f3(x8:0, x9:0 - 1) :|: x9:0 > -1 && x8:0 > -1
f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f3(x, x1)] = -1 + x

The following rules are decreasing:
f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0
The following rules are bounded:
f3(x6:0, x7:0) -> f3(x6:0 - 1, x5:0 - 1) :|: x6:0 > 0 && x5:0 > -1 && x7:0 < 0

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(8)
Obligation:
Rules:
f3(x8:0, x9:0) -> f3(x8:0, x9:0 - 1) :|: x9:0 > -1 && x8:0 > -1

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f3(x, x1)] = x1

The following rules are decreasing:
f3(x8:0, x9:0) -> f3(x8:0, x9:0 - 1) :|: x9:0 > -1 && x8:0 > -1
The following rules are bounded:
f3(x8:0, x9:0) -> f3(x8:0, x9:0 - 1) :|: x9:0 > -1 && x8:0 > -1

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(10)
YES