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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(s(0))
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f(s(0)) → *(s(s(0)), f(0))
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f(+(x, s(0))) → +(s(s(0)), f(x))
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f(+(x, y)) → *(f(x), f(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:
f(0) → s(0)
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f(s(0)) → s(s(0))
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f(s(0)) → *(s(s(0)), f(0))
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f(+(z0, s(0))) → +(s(s(0)), f(z0))
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f(+(z0, z1)) → *(f(z0), f(z1))
Tuples:
F(s(0)) → c2(F(0))
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F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
S tuples:
F(s(0)) → c2(F(0))
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F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2, c3, c4
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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) → s(0)
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f(s(0)) → s(s(0))
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f(s(0)) → *(s(s(0)), f(0))
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f(+(z0, s(0))) → +(s(s(0)), f(z0))
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f(+(z0, z1)) → *(f(z0), f(z1))
Tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
S tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c3, c4, c2
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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
F(s(0)) → c2
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → s(0)
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f(s(0)) → s(s(0))
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f(s(0)) → *(s(s(0)), f(0))
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f(+(z0, s(0))) → +(s(s(0)), f(z0))
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f(+(z0, z1)) → *(f(z0), f(z1))
Tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
S tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c3, c4, c2
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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.
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
We considered the (Usable) Rules:none
And the Tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
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(0) = 0
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POL(F(x1)) = [4]x1
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POL(c2) = 0
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POL(c3(x1)) = x1
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POL(c4(x1, x2)) = x1 + x2
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POL(s(x1)) = 0
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(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → s(0)
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f(s(0)) → s(s(0))
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f(s(0)) → *(s(s(0)), f(0))
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f(+(z0, s(0))) → +(s(s(0)), f(z0))
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f(+(z0, z1)) → *(f(z0), f(z1))
Tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
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F(s(0)) → c2
S tuples:
F(s(0)) → c2
K tuples:
F(+(z0, s(0))) → c3(F(z0))
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F(+(z0, z1)) → c4(F(z0), F(z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c3, c4, c2
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(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(s(0)) → c2
Now S is empty
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(10) BOUNDS(O(1), O(1))
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