/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 59 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) CaseAnalysis [EQUIVALENT, 20 ms]
(8) AND
    (9) IntTRS
        (10) TerminationGraphProcessor [EQUIVALENT, 17 ms]
        (11) IntTRS
        (12) IntTRSCompressionProof [EQUIVALENT, 4 ms]
        (13) IntTRS
        (14) RankingReductionPairProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) IntTRS
        (17) TerminationGraphProcessor [EQUIVALENT, 12 ms]
        (18) IntTRS
        (19) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (20) IntTRS
        (21) PolynomialOrderProcessor [EQUIVALENT, 1 ms]
        (22) 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:
f1(x) -> f2(x_1) :|: TRUE
f4(x1) -> f7(arith) :|: TRUE && arith = x1 - 1
f5(x10) -> f8(x11) :|: TRUE && x11 = x10 + 1
f3(x3) -> f4(x3) :|: x3 > 0
f3(x4) -> f5(x4) :|: x4 <= 0
f7(x5) -> f6(x5) :|: TRUE
f8(x6) -> f6(x6) :|: TRUE
f2(x7) -> f3(x7) :|: x7 < 0
f2(x12) -> f3(x12) :|: x12 > 0
f6(x8) -> f2(x8) :|: TRUE
f2(x9) -> f9(x9) :|: x9 = 0
Start term: f1(x)

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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(x7) -> f3(x7) :|: x7 < 0
f6(x8) -> f2(x8) :|: TRUE
f7(x5) -> f6(x5) :|: TRUE
f4(x1) -> f7(arith) :|: TRUE && arith = x1 - 1
f3(x3) -> f4(x3) :|: x3 > 0
f2(x12) -> f3(x12) :|: x12 > 0
f8(x6) -> f6(x6) :|: TRUE
f5(x10) -> f8(x11) :|: TRUE && x11 = x10 + 1
f3(x4) -> f5(x4) :|: x4 <= 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x8:0) -> f3(x8:0) :|: x8:0 < 0
f6(x) -> f3(x) :|: x > 0
f3(x4:0) -> f6(x4:0 + 1) :|: x4:0 < 1
f3(x3:0) -> f6(x3:0 - 1) :|: x3:0 > 0

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(7) CaseAnalysis (EQUIVALENT)
Found the following inductive condition: 
f6(x): 3 - 3*x>=0
f3(x1): 3 - 3*x1>=0

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(8)
Complex Obligation (AND)

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(9)
Obligation:
Rules:
f6(x8:0) -> f3(x8:0) :|: x8:0 < 0 && 3 + -3 * x8:0 >= 0
f6(x) -> f3(x) :|: x > 0 && 3 + -3 * x >= 0
f3(x4:0) -> f6(x4:0 + 1) :|: x4:0 < 1 && 3 + -3 * x4:0 >= 0
f3(x3:0) -> f6(x3:0 - 1) :|: x3:0 > 0 && 3 + -3 * x3:0 >= 0

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

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(11)
Obligation:
Rules:
f6(x8:0) -> f3(x8:0) :|: x8:0 < 0 && 3 + -3 * x8:0 >= 0
f3(x4:0) -> f6(x4:0 + 1) :|: x4:0 < 1 && 3 + -3 * x4:0 >= 0
f6(x) -> f3(x) :|: x > 0 && 3 + -3 * x >= 0

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(12) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(13)
Obligation:
Rules:
f6(x:0) -> f6(x:0 + 1) :|: x:0 > 0 && -3 * x:0 >= -3 && x:0 < 1
f6(x8:0:0) -> f6(x8:0:0 + 1) :|: x8:0:0 < 0 && -3 * x8:0:0 >= -3 && x8:0:0 < 1

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(14) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f6 ] = -1*f6_1

The following rules are decreasing:
f6(x:0) -> f6(x:0 + 1) :|: x:0 > 0 && -3 * x:0 >= -3 && x:0 < 1
f6(x8:0:0) -> f6(x8:0:0 + 1) :|: x8:0:0 < 0 && -3 * x8:0:0 >= -3 && x8:0:0 < 1

The following rules are bounded:
f6(x:0) -> f6(x:0 + 1) :|: x:0 > 0 && -3 * x:0 >= -3 && x:0 < 1
f6(x8:0:0) -> f6(x8:0:0 + 1) :|: x8:0:0 < 0 && -3 * x8:0:0 >= -3 && x8:0:0 < 1


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(15)
YES

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(16)
Obligation:
Rules:
f6(x8:0) -> f3(x8:0) :|: x8:0 < 0 && 3 + -3 * x8:0 < 0
f6(x) -> f3(x) :|: x > 0 && 3 + -3 * x < 0
f3(x4:0) -> f6(x4:0 + 1) :|: x4:0 < 1 && 3 + -3 * x4:0 < 0
f3(x3:0) -> f6(x3:0 - 1) :|: x3:0 > 0 && 3 + -3 * x3:0 < 0

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

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(18)
Obligation:
Rules:
f6(x) -> f3(x) :|: x > 0 && 3 + -3 * x < 0
f3(x3:0) -> f6(x3:0 - 1) :|: x3:0 > 0 && 3 + -3 * x3:0 < 0

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(19) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(20)
Obligation:
Rules:
f6(x:0) -> f6(x:0 - 1) :|: -3 * x:0 < -3 && x:0 > 0

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(21) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x)] = x

The following rules are decreasing:
f6(x:0) -> f6(x:0 - 1) :|: -3 * x:0 < -3 && x:0 > 0
The following rules are bounded:
f6(x:0) -> f6(x:0 - 1) :|: -3 * x:0 < -3 && x:0 > 0

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(22)
YES