YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.c
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination of the given C Problem could be proven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) TerminationGraphProcessor [SOUND, 81 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 32 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(8) IntTRS
(9) TerminationGraphProcessor [EQUIVALENT, 14 ms]
(10) AND
    (11) IntTRS
        (12) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
        (13) IntTRS
        (14) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
        (15) YES
    (16) IntTRS
        (17) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (18) IntTRS
        (19) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
        (20) AND
            (21) IntTRS
                (22) TerminationGraphProcessor [EQUIVALENT, 1 ms]
                (23) YES
            (24) IntTRS
                (25) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
                (26) IntTRS
                (27) RankingReductionPairProof [EQUIVALENT, 3 ms]
                (28) 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, y) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f8(x4, x5) -> f9(arith, x5) :|: TRUE && arith = x4 - 1
f5(x6, x7) -> f8(x6, x7) :|: x6 > 0
f9(x8, x9) -> f5(x8, x9) :|: TRUE
f5(x10, x11) -> f10(x10, x11) :|: x10 <= 0
f11(x34, x35) -> f12(x34, x36) :|: TRUE && x36 = x35 - 1
f6(x14, x15) -> f11(x14, x15) :|: x15 > 0
f12(x16, x17) -> f6(x16, x17) :|: TRUE
f6(x18, x19) -> f13(x18, x19) :|: x19 <= 0
f4(x20, x21) -> f5(x20, x21) :|: x20 > x21
f4(x22, x23) -> f6(x22, x23) :|: x22 <= x23
f10(x24, x25) -> f7(x24, x25) :|: TRUE
f13(x26, x27) -> f7(x26, x27) :|: TRUE
f3(x28, x29) -> f4(x28, x29) :|: x28 > 0 && x29 > 0
f7(x30, x31) -> f3(x30, x31) :|: TRUE
f3(x32, x33) -> f14(x32, x33) :|: x32 <= 0
f3(x37, x38) -> f14(x37, x38) :|: x38 <= 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:
f3(x28, x29) -> f4(x28, x29) :|: x28 > 0 && x29 > 0
f7(x30, x31) -> f3(x30, x31) :|: TRUE
f10(x24, x25) -> f7(x24, x25) :|: TRUE
f5(x10, x11) -> f10(x10, x11) :|: x10 <= 0
f4(x20, x21) -> f5(x20, x21) :|: x20 > x21
f9(x8, x9) -> f5(x8, x9) :|: TRUE
f8(x4, x5) -> f9(arith, x5) :|: TRUE && arith = x4 - 1
f5(x6, x7) -> f8(x6, x7) :|: x6 > 0
f13(x26, x27) -> f7(x26, x27) :|: TRUE
f6(x18, x19) -> f13(x18, x19) :|: x19 <= 0
f4(x22, x23) -> f6(x22, x23) :|: x22 <= x23
f12(x16, x17) -> f6(x16, x17) :|: TRUE
f11(x34, x35) -> f12(x34, x36) :|: TRUE && x36 = x35 - 1
f6(x14, x15) -> f11(x14, x15) :|: x15 > 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x6:0, x7:0) -> f5(x6:0 - 1, x7:0) :|: x6:0 > 0
f7(x30:0, x31:0) -> f6(x30:0, x31:0) :|: x30:0 > 0 && x31:0 > 0 && x31:0 >= x30:0
f7(x, x1) -> f5(x, x1) :|: x > 0 && x1 > 0 && x1 < x
f5(x10:0, x11:0) -> f7(x10:0, x11:0) :|: x10:0 < 1
f6(x14:0, x15:0) -> f6(x14:0, x15:0 - 1) :|: x15:0 > 0
f6(x18:0, x19:0) -> f7(x18:0, x19:0) :|: x19:0 < 1

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = x1
[f7(x2, x3)] = x3
[f6(x4, x5)] = 0

The following rules are decreasing:
f7(x30:0, x31:0) -> f6(x30:0, x31:0) :|: x30:0 > 0 && x31:0 > 0 && x31:0 >= x30:0
The following rules are bounded:
f7(x30:0, x31:0) -> f6(x30:0, x31:0) :|: x30:0 > 0 && x31:0 > 0 && x31:0 >= x30:0
f6(x14:0, x15:0) -> f6(x14:0, x15:0 - 1) :|: x15:0 > 0
f6(x18:0, x19:0) -> f7(x18:0, x19:0) :|: x19:0 < 1

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(8)
Obligation:
Rules:
f5(x6:0, x7:0) -> f5(x6:0 - 1, x7:0) :|: x6:0 > 0
f7(x, x1) -> f5(x, x1) :|: x > 0 && x1 > 0 && x1 < x
f5(x10:0, x11:0) -> f7(x10:0, x11:0) :|: x10:0 < 1
f6(x14:0, x15:0) -> f6(x14:0, x15:0 - 1) :|: x15:0 > 0
f6(x18:0, x19:0) -> f7(x18:0, x19:0) :|: x19:0 < 1

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(9) TerminationGraphProcessor (EQUIVALENT)
Constructed the termination graph and obtained 2 non-trivial SCCs.

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(10)
Complex Obligation (AND)

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(11)
Obligation:
Rules:
f6(x14:0, x15:0) -> f6(x14:0, x15:0 - 1) :|: x15:0 > 0

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(12) IntTRSUnneededArgumentFilterProof (EQUIVALENT)
Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements:

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

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(13)
Obligation:
Rules:
f6(x15:0) -> f6(x15:0 - 1) :|: x15:0 > 0

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

The following rules are decreasing:
f6(x15:0) -> f6(x15:0 - 1) :|: x15:0 > 0
The following rules are bounded:
f6(x15:0) -> f6(x15:0 - 1) :|: x15:0 > 0

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(15)
YES

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(16)
Obligation:
Rules:
f5(x6:0, x7:0) -> f5(x6:0 - 1, x7:0) :|: x6:0 > 0
f7(x, x1) -> f5(x, x1) :|: x > 0 && x1 > 0 && x1 < x
f5(x10:0, x11:0) -> f7(x10:0, x11:0) :|: x10:0 < 1

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(17) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(18)
Obligation:
Rules:
f5(x6:0:0, x7:0:0) -> f5(x6:0:0 - 1, x7:0:0) :|: x6:0:0 > 0
f5(x10:0:0, x11:0:0) -> f5(x10:0:0, x11:0:0) :|: x11:0:0 < x10:0:0 && x10:0:0 < 1 && x11:0:0 > 0 && x10:0:0 > 0

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(19) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = d + x + c1*x1

The following rules are decreasing:
f5(x6:0:0, x7:0:0) -> f5(x6:0:0 - 1, x7:0:0) :|: x6:0:0 > 0
The following rules are bounded:
f5(x10:0:0, x11:0:0) -> f5(x10:0:0, x11:0:0) :|: x11:0:0 < x10:0:0 && x10:0:0 < 1 && x11:0:0 > 0 && x10:0:0 > 0

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(20)
Complex Obligation (AND)

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(21)
Obligation:
Rules:
f5(x10:0:0, x11:0:0) -> f5(x10:0:0, x11:0:0) :|: x11:0:0 < x10:0:0 && x10:0:0 < 1 && x11:0:0 > 0 && x10:0:0 > 0

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(22) TerminationGraphProcessor (EQUIVALENT)
Constructed the termination graph and obtained no non-trivial SCC(s).

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(23)
YES

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(24)
Obligation:
Rules:
f5(x6:0:0, x7:0:0) -> f5(x6:0:0 - 1, x7:0:0) :|: x6:0:0 > 0

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(25) IntTRSUnneededArgumentFilterProof (EQUIVALENT)
Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements:

   f5(x1, x2) -> f5(x1)

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(26)
Obligation:
Rules:
f5(x6:0:0) -> f5(x6:0:0 - 1) :|: x6:0:0 > 0

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(27) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = f5_1

The following rules are decreasing:
f5(x6:0:0) -> f5(x6:0:0 - 1) :|: x6:0:0 > 0

The following rules are bounded:
f5(x6:0:0) -> f5(x6:0:0 - 1) :|: x6:0:0 > 0


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(28)
YES