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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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), 10 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)), 27 ms]
    (8) CdtProblem
    (9) CdtKnowledgeProof [FINISHED, 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:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

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:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))
Tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:   quot_3

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c, c1_1, c2_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   QUOT(0, s(z0), s(z1)) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:   quot_3

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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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.
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(QUOT(x_1, x_2, x_3)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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(9) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
Now 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:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

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

quot(s(x), s(y), z) ->^+ quot(x, y, z)

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

The pumping substitution is [x / s(x), y / s(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:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

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:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

S is empty.
Rewrite Strategy: INNERMOST