YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 47 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) CaseAnalysis [EQUIVALENT, 69 ms]
(10) AND
    (11) IntTRS
        (12) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (13) IntTRS
        (14) PolynomialOrderProcessor [EQUIVALENT, 7 ms]
        (15) YES
    (16) IntTRS
        (17) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (18) IntTRS
        (19) RankingReductionPairProof [EQUIVALENT, 0 ms]
        (20) 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, oldx) -> f2(x_1, y, oldx) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1
f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1
f3(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0
f3(x27, x28, x29) -> f4(x27, x28, x29) :|: x28 >= 0
f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE
f3(x20, x21, x22) -> f8(x20, x21, x22) :|: x20 < 0 && x21 < 0
Start term: f1(x, y, oldx)

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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(x14, x15, x16) -> f4(x14, x15, x16) :|: x14 >= 0
f7(x17, x18, x19) -> f3(x17, x18, x19) :|: TRUE
f6(x23, x24, x25) -> f7(x23, x26, x25) :|: TRUE && x26 = x25 - 1
f5(x8, x9, x10) -> f6(arith, x9, x10) :|: TRUE && arith = x9 - 1
f4(x5, x6, x7) -> f5(x5, x6, x5) :|: TRUE
f3(x27, x28, x29) -> f4(x27, x28, x29) :|: x28 >= 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x8:0, x9:0, x10:0) -> f5(x9:0 - 1, x10:0 - 1, x9:0 - 1) :|: x10:0 > 0
f5(x, x1, x2) -> f5(x1 - 1, x2 - 1, x1 - 1) :|: x1 > 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:

   f5(x1, x2, x3) -> f5(x2, x3)

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(8)
Obligation:
Rules:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x10:0 > 0
f5(x1, x2) -> f5(x2 - 1, x1 - 1) :|: x1 > 0

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(9) CaseAnalysis (EQUIVALENT)
Found the following inductive condition: 
f5(x, x1): -1 - 3*x - 3*x1>=0

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(10)
Complex Obligation (AND)

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(11)
Obligation:
Rules:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x10:0 > 0 && -1 + -3 * x9:0 + -3 * x10:0 >= 0
f5(x1, x2) -> f5(x2 - 1, x1 - 1) :|: x1 > 0 && -1 + -3 * x1 + -3 * x2 >= 0

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(12) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(13)
Obligation:
Rules:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 <= -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 <= -1 + -3 * x1:0 + -3 * x2:0

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(14) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = -2 - x*x1

The following rules are decreasing:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 <= -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 <= -1 + -3 * x1:0 + -3 * x2:0
The following rules are bounded:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 <= -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 <= -1 + -3 * x1:0 + -3 * x2:0

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(15)
YES

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(16)
Obligation:
Rules:
f5(x9:0, x10:0) -> f5(x10:0 - 1, x9:0 - 1) :|: x10:0 > 0 && -1 + -3 * x9:0 + -3 * x10:0 < 0
f5(x1, x2) -> f5(x2 - 1, x1 - 1) :|: x1 > 0 && -1 + -3 * x1 + -3 * x2 < 0

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(17) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(18)
Obligation:
Rules:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 > -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 > -1 + -3 * x1:0 + -3 * x2:0

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(19) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = 1/2*f5_2 + 1/2*f5_1

The following rules are decreasing:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 > -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 > -1 + -3 * x1:0 + -3 * x2:0

The following rules are bounded:
f5(x9:0:0, x10:0:0) -> f5(x10:0:0 - 1, x9:0:0 - 1) :|: x10:0:0 > 0 && 0 > -1 + -3 * x9:0:0 + -3 * x10:0:0
f5(x1:0, x2:0) -> f5(x2:0 - 1, x1:0 - 1) :|: x1:0 > 0 && 0 > -1 + -3 * x1:0 + -3 * x2:0


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(20)
YES