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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
+(0, y) → y
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+(s(x), y) → s(+(x, y))
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+(s(x), y) → +(x, s(y))
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:
+(0, z0) → z0
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+(s(z0), z1) → s(+(z0, z1))
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+(s(z0), z1) → +(z0, s(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
S tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
K tuples:none
Defined Rule Symbols:
+
Defined Pair Symbols:
+'
Compound Symbols:
c1, 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.
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
We considered the (Usable) Rules:none
And the Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(+'(x1, x2)) = [2]x12
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POL(c1(x1)) = x1
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POL(c2(x1)) = x1
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POL(s(x1)) = [2] + x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
+(0, z0) → z0
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+(s(z0), z1) → s(+(z0, z1))
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+(s(z0), z1) → +(z0, s(z1))
Tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
S tuples:none
K tuples:
+'(s(z0), z1) → c1(+'(z0, z1))
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+'(s(z0), z1) → c2(+'(z0, s(z1)))
Defined Rule Symbols:
+
Defined Pair Symbols:
+'
Compound Symbols:
c1, 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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