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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
rev(a) → a
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rev(b) → b
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rev(++(x, y)) → ++(rev(y), rev(x))
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rev(++(x, x)) → rev(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:
rev(a) → a
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rev(b) → b
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rev(++(z0, z1)) → ++(rev(z1), rev(z0))
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rev(++(z0, z0)) → rev(z0)
Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
S tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:
rev
Defined Pair Symbols:
REV
Compound Symbols:
c2, c3
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(3) 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.
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
We considered the (Usable) Rules:none
And the Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(++(x1, x2)) = [1] + x1 + x2
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POL(REV(x1)) = [2]x1
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POL(c2(x1, x2)) = x1 + x2
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POL(c3(x1)) = x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
rev(a) → a
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rev(b) → b
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rev(++(z0, z1)) → ++(rev(z1), rev(z0))
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rev(++(z0, z0)) → rev(z0)
Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
S tuples:none
K tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
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REV(++(z0, z0)) → c3(REV(z0))
Defined Rule Symbols:
rev
Defined Pair Symbols:
REV
Compound Symbols:
c2, c3
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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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