/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 47 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 38 ms]
(6) IntTRS
(7) CaseAnalysis [EQUIVALENT, 21 ms]
(8) AND
    (9) IntTRS
        (10) TerminationGraphProcessor [EQUIVALENT, 18 ms]
        (11) IntTRS
        (12) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (13) IntTRS
        (14) RankingReductionPairProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) IntTRS
        (17) TerminationGraphProcessor [EQUIVALENT, 17 ms]
        (18) IntTRS
        (19) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (20) IntTRS
        (21) PolynomialOrderProcessor [EQUIVALENT, 0 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(K, x) -> f2(x_1, x) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f5(x4, x5) -> f8(x4, arith) :|: TRUE && arith = x5 - 1
f6(x22, x23) -> f9(x22, x24) :|: TRUE && x24 = x23 + 1
f4(x8, x9) -> f5(x8, x9) :|: x9 > x8
f4(x10, x11) -> f6(x10, x11) :|: x11 <= x10
f8(x12, x13) -> f7(x12, x13) :|: TRUE
f9(x14, x15) -> f7(x14, x15) :|: TRUE
f3(x16, x17) -> f4(x16, x17) :|: x17 < x16
f3(x25, x26) -> f4(x25, x26) :|: x26 > x25
f7(x18, x19) -> f3(x18, x19) :|: TRUE
f3(x20, x21) -> f10(x20, x21) :|: x21 = x20
Start term: f1(K, 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:
f3(x16, x17) -> f4(x16, x17) :|: x17 < x16
f7(x18, x19) -> f3(x18, x19) :|: TRUE
f8(x12, x13) -> f7(x12, x13) :|: TRUE
f5(x4, x5) -> f8(x4, arith) :|: TRUE && arith = x5 - 1
f4(x8, x9) -> f5(x8, x9) :|: x9 > x8
f3(x25, x26) -> f4(x25, x26) :|: x26 > x25
f9(x14, x15) -> f7(x14, x15) :|: TRUE
f6(x22, x23) -> f9(x22, x24) :|: TRUE && x24 = x23 + 1
f4(x10, x11) -> f6(x10, x11) :|: x11 <= x10

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f7(x18:0, x19:0) -> f4(x18:0, x19:0) :|: x19:0 > x18:0
f4(x10:0, x11:0) -> f7(x10:0, x11:0 + 1) :|: x11:0 <= x10:0
f4(x8:0, x9:0) -> f7(x8:0, x9:0 - 1) :|: x9:0 > x8:0
f7(x, x1) -> f4(x, x1) :|: x1 < x

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(7) CaseAnalysis (EQUIVALENT)
Found the following inductive condition: 
f7(x, x1): -3*x + 3*x1>=0
f4(x2, x3): -3*x2 + 3*x3>=0

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(8)
Complex Obligation (AND)

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(9)
Obligation:
Rules:
f7(x18:0, x19:0) -> f4(x18:0, x19:0) :|: x19:0 > x18:0 && -3 * x18:0 + 3 * x19:0 >= 0
f4(x10:0, x11:0) -> f7(x10:0, x11:0 + 1) :|: x11:0 <= x10:0 && -3 * x10:0 + 3 * x11:0 >= 0
f4(x8:0, x9:0) -> f7(x8:0, x9:0 - 1) :|: x9:0 > x8:0 && -3 * x8:0 + 3 * x9:0 >= 0
f7(x, x1) -> f4(x, x1) :|: x1 < x && -3 * x + 3 * x1 >= 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:
f7(x18:0, x19:0) -> f4(x18:0, x19:0) :|: x19:0 > x18:0 && -3 * x18:0 + 3 * x19:0 >= 0
f4(x10:0, x11:0) -> f7(x10:0, x11:0 + 1) :|: x11:0 <= x10:0 && -3 * x10:0 + 3 * x11:0 >= 0
f4(x8:0, x9:0) -> f7(x8:0, x9:0 - 1) :|: x9:0 > x8:0 && -3 * x8:0 + 3 * x9:0 >= 0

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(12) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(13)
Obligation:
Rules:
f7(x18:0:0, x19:0:0) -> f7(x18:0:0, x19:0:0 - 1) :|: 0 <= -3 * x18:0:0 + 3 * x19:0:0 && x19:0:0 > x18:0:0

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(14) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f7 ] = f7_2 + -1*f7_1

The following rules are decreasing:
f7(x18:0:0, x19:0:0) -> f7(x18:0:0, x19:0:0 - 1) :|: 0 <= -3 * x18:0:0 + 3 * x19:0:0 && x19:0:0 > x18:0:0

The following rules are bounded:
f7(x18:0:0, x19:0:0) -> f7(x18:0:0, x19:0:0 - 1) :|: 0 <= -3 * x18:0:0 + 3 * x19:0:0 && x19:0:0 > x18:0:0


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(15)
YES

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(16)
Obligation:
Rules:
f7(x18:0, x19:0) -> f4(x18:0, x19:0) :|: x19:0 > x18:0 && -3 * x18:0 + 3 * x19:0 < 0
f4(x10:0, x11:0) -> f7(x10:0, x11:0 + 1) :|: x11:0 <= x10:0 && -3 * x10:0 + 3 * x11:0 < 0
f4(x8:0, x9:0) -> f7(x8:0, x9:0 - 1) :|: x9:0 > x8:0 && -3 * x8:0 + 3 * x9:0 < 0
f7(x, x1) -> f4(x, x1) :|: x1 < x && -3 * x + 3 * x1 < 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:
f4(x10:0, x11:0) -> f7(x10:0, x11:0 + 1) :|: x11:0 <= x10:0 && -3 * x10:0 + 3 * x11:0 < 0
f7(x, x1) -> f4(x, x1) :|: x1 < x && -3 * x + 3 * x1 < 0

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(19) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(20)
Obligation:
Rules:
f4(x10:0:0, x11:0:0) -> f4(x10:0:0, x11:0:0 + 1) :|: 0 > -3 * x10:0:0 + 3 * x11:0:0 && x11:0:0 <= x10:0:0 && 0 > -3 * x10:0:0 + 3 * (x11:0:0 + 1) && x11:0:0 + 1 < x10:0:0

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(21) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f4(x, x1)] = x - x1

The following rules are decreasing:
f4(x10:0:0, x11:0:0) -> f4(x10:0:0, x11:0:0 + 1) :|: 0 > -3 * x10:0:0 + 3 * x11:0:0 && x11:0:0 <= x10:0:0 && 0 > -3 * x10:0:0 + 3 * (x11:0:0 + 1) && x11:0:0 + 1 < x10:0:0
The following rules are bounded:
f4(x10:0:0, x11:0:0) -> f4(x10:0:0, x11:0:0 + 1) :|: 0 > -3 * x10:0:0 + 3 * x11:0:0 && x11:0:0 <= x10:0:0 && 0 > -3 * x10:0:0 + 3 * (x11:0:0 + 1) && x11:0:0 + 1 < x10:0:0

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(22)
YES