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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(X) → g(n__h(n__f(X)))
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h(X) → n__h(X)
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f(X) → n__f(X)
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activate(n__h(X)) → h(activate(X))
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activate(n__f(X)) → f(activate(X))
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activate(X) → X
Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
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(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0) → g(n__h(n__f(z0)))
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f(z0) → n__f(z0)
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h(z0) → n__h(z0)
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activate(n__h(z0)) → h(activate(z0))
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activate(n__f(z0)) → f(activate(z0))
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activate(z0) → z0
Tuples:
ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
S tuples:
ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, h, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3, c4
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(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0) → g(n__h(n__f(z0)))
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f(z0) → n__f(z0)
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h(z0) → n__h(z0)
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activate(n__h(z0)) → h(activate(z0))
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activate(n__f(z0)) → f(activate(z0))
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activate(z0) → z0
Tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, h, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3, c4
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(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(ACTIVATE(x1)) = [2]x1
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POL(c3(x1)) = x1
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POL(c4(x1)) = x1
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POL(n__f(x1)) = x1
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POL(n__h(x1)) = [1] + x1
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0) → g(n__h(n__f(z0)))
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f(z0) → n__f(z0)
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h(z0) → n__h(z0)
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activate(n__h(z0)) → h(activate(z0))
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activate(n__f(z0)) → f(activate(z0))
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activate(z0) → z0
Tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
K tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
Defined Rule Symbols:
f, h, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3, c4
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(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(ACTIVATE(x1)) = [3]x1
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POL(c3(x1)) = x1
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POL(c4(x1)) = x1
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POL(n__f(x1)) = [1] + x1
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POL(n__h(x1)) = x1
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(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0) → g(n__h(n__f(z0)))
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f(z0) → n__f(z0)
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h(z0) → n__h(z0)
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activate(n__h(z0)) → h(activate(z0))
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activate(n__f(z0)) → f(activate(z0))
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activate(z0) → z0
Tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:none
K tuples:
ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
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ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
Defined Rule Symbols:
f, h, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3, c4
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(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
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(10) BOUNDS(O(1), O(1))
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