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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0) → s(0)
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f(s(0)) → s(0)
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f(s(s(x))) → f(f(s(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(0) → s(0)
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f(s(0)) → s(0)
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f(s(s(z0))) → f(f(s(z0)))
Tuples:
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
S tuples:
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2
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(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
We considered the (Usable) Rules:
f(s(0)) → s(0)
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f(s(s(z0))) → f(f(s(z0)))
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f(0) → s(0)
And the Tuples:
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = [1]
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POL(F(x1)) = [1] + [3]x12
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POL(c2(x1, x2)) = x1 + x2
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POL(f(x1)) = [3]
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POL(s(x1)) = [2] + x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → s(0)
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f(s(0)) → s(0)
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f(s(s(z0))) → f(f(s(z0)))
Tuples:
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
S tuples:none
K tuples:
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2
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(5) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
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(6) BOUNDS(O(1), O(1))
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