WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 11 ms]
(2) CdtProblem
    (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CdtProblem
    (5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 44 ms]
    (8) CdtProblem
    (9) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (10) BOUNDS(1, 1)
(11) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(12) TRS for Loop Detection
    (13) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (14) BEST
        (15) proven lower bound
            (16) LowerBoundPropagationProof [FINISHED, 0 ms]
            (17) BOUNDS(n^1, INF)
        (18) TRS for Loop Detection


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(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   D(t) -> 1
   D(constant) -> 0
   D(+(x, y)) -> +(D(x), D(y))
   D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
   D(-(x, y)) -> -(D(x), D(y))

S is empty.
Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   D(t) -> 1
   D(constant) -> 0
   D(+(z0, z1)) -> +(D(z0), D(z1))
   D(*(z0, z1)) -> +(*(z1, D(z0)), *(z0, D(z1)))
   D(-(z0, z1)) -> -(D(z0), D(z1))
Tuples:
   D'(t) -> c
   D'(constant) -> c1
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
S tuples:
   D'(t) -> c
   D'(constant) -> c1
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:   D_1

Defined Pair Symbols:   D'_1

Compound Symbols:   c, c1, c2_2, c3_2, c4_2


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   D'(t) -> c
   D'(constant) -> c1

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   D(t) -> 1
   D(constant) -> 0
   D(+(z0, z1)) -> +(D(z0), D(z1))
   D(*(z0, z1)) -> +(*(z1, D(z0)), *(z0, D(z1)))
   D(-(z0, z1)) -> -(D(z0), D(z1))
Tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
S tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:   D_1

Defined Pair Symbols:   D'_1

Compound Symbols:   c2_2, c3_2, c4_2


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   D(t) -> 1
   D(constant) -> 0
   D(+(z0, z1)) -> +(D(z0), D(z1))
   D(*(z0, z1)) -> +(*(z1, D(z0)), *(z0, D(z1)))
   D(-(z0, z1)) -> -(D(z0), D(z1))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
S tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   D'_1

Compound Symbols:   c2_2, c3_2, c4_2


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(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
We considered the (Usable) Rules:none
And the Tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(*(x_1, x_2)) = [1] + x_1 + x_2
   POL(+(x_1, x_2)) = [1] + x_1 + x_2
   POL(-(x_1, x_2)) = [1] + x_1 + x_2
   POL(D'(x_1)) = x_1
   POL(c2(x_1, x_2)) = x_1 + x_2
   POL(c3(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1, x_2)) = x_1 + x_2

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
S tuples:none
K tuples:
   D'(+(z0, z1)) -> c2(D'(z0), D'(z1))
   D'(*(z0, z1)) -> c3(D'(z0), D'(z1))
   D'(-(z0, z1)) -> c4(D'(z0), D'(z1))
Defined Rule Symbols:none

Defined Pair Symbols:   D'_1

Compound Symbols:   c2_2, c3_2, c4_2


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(9) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(10)
BOUNDS(1, 1)

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(11) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
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(12)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   D(t) -> 1
   D(constant) -> 0
   D(+(x, y)) -> +(D(x), D(y))
   D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
   D(-(x, y)) -> -(D(x), D(y))

S is empty.
Rewrite Strategy: INNERMOST
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(13) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

D(+(x, y)) ->^+ +(D(x), D(y))

gives rise to a decreasing loop by considering the right hand sides subterm at position [0].

The pumping substitution is [x / +(x, y)].

The result substitution is [ ].




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(14)
Complex Obligation (BEST)

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(15)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   D(t) -> 1
   D(constant) -> 0
   D(+(x, y)) -> +(D(x), D(y))
   D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
   D(-(x, y)) -> -(D(x), D(y))

S is empty.
Rewrite Strategy: INNERMOST
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(16) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(17)
BOUNDS(n^1, INF)

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(18)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   D(t) -> 1
   D(constant) -> 0
   D(+(x, y)) -> +(D(x), D(y))
   D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
   D(-(x, y)) -> -(D(x), D(y))

S is empty.
Rewrite Strategy: INNERMOST