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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(a) → f(c(a))
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f(c(X)) → X
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f(c(a)) → f(d(b))
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f(a) → f(d(a))
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f(d(X)) → X
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f(c(b)) → f(d(a))
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e(g(X)) → e(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:
f(a) → f(c(a))
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f(c(z0)) → z0
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f(c(a)) → f(d(b))
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f(a) → f(d(a))
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f(d(z0)) → z0
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f(c(b)) → f(d(a))
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e(g(z0)) → e(z0)
Tuples:
F(a) → c1(F(c(a)))
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F(c(a)) → c3(F(d(b)))
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F(a) → c4(F(d(a)))
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F(c(b)) → c6(F(d(a)))
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E(g(z0)) → c7(E(z0))
S tuples:
F(a) → c1(F(c(a)))
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F(c(a)) → c3(F(d(b)))
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F(a) → c4(F(d(a)))
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F(c(b)) → c6(F(d(a)))
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E(g(z0)) → c7(E(z0))
K tuples:none
Defined Rule Symbols:
f, e
Defined Pair Symbols:
F, E
Compound Symbols:
c1, c3, c4, c6, c7
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(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing tuple parts
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → f(c(a))
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f(c(z0)) → z0
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f(c(a)) → f(d(b))
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f(a) → f(d(a))
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f(d(z0)) → z0
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f(c(b)) → f(d(a))
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e(g(z0)) → e(z0)
Tuples:
F(a) → c1(F(c(a)))
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E(g(z0)) → c7(E(z0))
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F(c(a)) → c3
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F(a) → c4
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F(c(b)) → c6
S tuples:
F(a) → c1(F(c(a)))
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E(g(z0)) → c7(E(z0))
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F(c(a)) → c3
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F(a) → c4
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F(c(b)) → c6
K tuples:none
Defined Rule Symbols:
f, e
Defined Pair Symbols:
F, E
Compound Symbols:
c1, c7, c3, c4, c6
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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
F(c(b)) → c6
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F(c(a)) → c3
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F(a) → c4
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F(a) → c1(F(c(a)))
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → f(c(a))
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f(c(z0)) → z0
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f(c(a)) → f(d(b))
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f(a) → f(d(a))
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f(d(z0)) → z0
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f(c(b)) → f(d(a))
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e(g(z0)) → e(z0)
Tuples:
E(g(z0)) → c7(E(z0))
S tuples:
E(g(z0)) → c7(E(z0))
K tuples:none
Defined Rule Symbols:
f, e
Defined Pair Symbols:
E
Compound Symbols:
c7
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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.
E(g(z0)) → c7(E(z0))
We considered the (Usable) Rules:none
And the Tuples:
E(g(z0)) → c7(E(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(E(x1)) = [3]x1
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POL(c7(x1)) = x1
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POL(g(x1)) = [1] + x1
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(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → f(c(a))
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f(c(z0)) → z0
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f(c(a)) → f(d(b))
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f(a) → f(d(a))
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f(d(z0)) → z0
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f(c(b)) → f(d(a))
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e(g(z0)) → e(z0)
Tuples:
E(g(z0)) → c7(E(z0))
S tuples:none
K tuples:
E(g(z0)) → c7(E(z0))
Defined Rule Symbols:
f, e
Defined Pair Symbols:
E
Compound Symbols:
c7
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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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