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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
g(x, y) → x
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g(x, y) → y
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f(0, 1, x) → f(s(x), x, x)
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f(x, y, s(z)) → s(f(0, 1, z))
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:
g(z0, z1) → z0
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g(z0, z1) → z1
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f(0, 1, z0) → f(s(z0), z0, z0)
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f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
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F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
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F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
F
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.
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
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F(z0, z1, s(z2)) → c3(F(0, 1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = [1]
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POL(1) = 0
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POL(F(x1, x2, x3)) = [4]x3
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POL(c2(x1)) = x1
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POL(c3(x1)) = x1
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POL(s(x1)) = [2] + x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(z0, z1) → z0
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g(z0, z1) → z1
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f(0, 1, z0) → f(s(z0), z0, z0)
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f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
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F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
K tuples:
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Defined Rule Symbols:
g, f
Defined Pair Symbols:
F
Compound Symbols:
c2, c3
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(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
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F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Now S is empty
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(6) BOUNDS(O(1), O(1))
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