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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0) → cons(0)
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f(s(0)) → f(p(s(0)))
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p(s(0)) → 0
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) → cons(0)
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f(s(0)) → f(p(s(0)))
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p(s(0)) → 0
Tuples:
F(s(0)) → c1(F(p(s(0))), P(s(0)))
S tuples:
F(s(0)) → c1(F(p(s(0))), P(s(0)))
K tuples:none
Defined Rule Symbols:
f, p
Defined Pair Symbols:
F
Compound Symbols:
c1
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(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0)
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f(s(0)) → f(p(s(0)))
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p(s(0)) → 0
Tuples:
F(s(0)) → c1(F(p(s(0))))
S tuples:
F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:
f, p
Defined Pair Symbols:
F
Compound Symbols:
c1
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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.
F(s(0)) → c1(F(p(s(0))))
We considered the (Usable) Rules:
p(s(0)) → 0
And the Tuples:
F(s(0)) → c1(F(p(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = [3]
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POL(F(x1)) = [4]x1
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POL(c1(x1)) = x1
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POL(p(x1)) = [4]
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POL(s(x1)) = [5] + x1
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0)
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f(s(0)) → f(p(s(0)))
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p(s(0)) → 0
Tuples:
F(s(0)) → c1(F(p(s(0))))
S tuples:none
K tuples:
F(s(0)) → c1(F(p(s(0))))
Defined Rule Symbols:
f, p
Defined Pair Symbols:
F
Compound Symbols:
c1
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(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
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(8) BOUNDS(O(1), O(1))
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