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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
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f(c(s(x), y)) → f(c(x, s(y)))
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f(f(x)) → f(d(f(x)))
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f(x) → 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:
g(c(z0, s(z1))) → g(c(s(z0), z1))
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f(c(s(z0), z1)) → f(c(z0, s(z1)))
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f(f(z0)) → f(d(f(z0)))
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f(z0) → z0
Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
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F(f(z0)) → c3(F(d(f(z0))), F(z0))
S tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
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F(f(z0)) → c3(F(d(f(z0))), F(z0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c1, c2, c3
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(3) CdtUnreachableProof (EQUIVALENT transformation)
The following tuples could be removed as they are not reachable from basic start terms:
F(f(z0)) → c3(F(d(f(z0))), F(z0))
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(c(z0, s(z1))) → g(c(s(z0), z1))
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f(c(s(z0), z1)) → f(c(z0, s(z1)))
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f(f(z0)) → f(d(f(z0)))
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f(z0) → z0
Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c1, c2
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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(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
We considered the (Usable) Rules:none
And the Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(F(x1)) = [4]x1
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POL(G(x1)) = 0
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POL(c(x1, x2)) = x1
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POL(c1(x1)) = x1
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POL(c2(x1)) = x1
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POL(s(x1)) = [4] + x1
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(c(z0, s(z1))) → g(c(s(z0), z1))
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f(c(s(z0), z1)) → f(c(z0, s(z1)))
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f(f(z0)) → f(d(f(z0)))
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f(z0) → z0
Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
K tuples:
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c1, 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.
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
We considered the (Usable) Rules:none
And the Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(F(x1)) = 0
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POL(G(x1)) = [4]x1
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POL(c(x1, x2)) = x2
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POL(c1(x1)) = x1
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POL(c2(x1)) = x1
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POL(s(x1)) = [4] + x1
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(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(c(z0, s(z1))) → g(c(s(z0), z1))
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f(c(s(z0), z1)) → f(c(z0, s(z1)))
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f(f(z0)) → f(d(f(z0)))
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f(z0) → z0
Tuples:
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
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F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:none
K tuples:
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
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G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c1, c2
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(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
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(10) BOUNDS(O(1), O(1))
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