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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
a__f(f(a)) → a__f(g(f(a)))
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mark(f(X)) → a__f(X)
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mark(a) → a
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mark(g(X)) → g(mark(X))
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a__f(X) → f(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:
a__f(f(a)) → a__f(g(f(a)))
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a__f(z0) → f(z0)
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mark(f(z0)) → a__f(z0)
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mark(a) → a
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mark(g(z0)) → g(mark(z0))
Tuples:
A__F(f(a)) → c(A__F(g(f(a))))
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MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
S tuples:
A__F(f(a)) → c(A__F(g(f(a))))
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MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
A__F, MARK
Compound Symbols:
c, c2, 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:
a__f(f(a)) → a__f(g(f(a)))
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a__f(z0) → f(z0)
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mark(f(z0)) → a__f(z0)
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mark(a) → a
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mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
S tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
K tuples:none
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK, A__F
Compound Symbols:
c2, c4, c
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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
A__F(f(a)) → c
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MARK(f(z0)) → c2(A__F(z0))
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(f(a)) → a__f(g(f(a)))
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a__f(z0) → f(z0)
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mark(f(z0)) → a__f(z0)
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mark(a) → a
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mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
S tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
K tuples:none
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK, A__F
Compound Symbols:
c2, c4, c
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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.
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(A__F(x1)) = 0
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POL(MARK(x1)) = [3] + [2]x1
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POL(a) = [1]
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POL(c) = 0
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POL(c2(x1)) = x1
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POL(c4(x1)) = x1
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POL(f(x1)) = [2]
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POL(g(x1)) = [3] + x1
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(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(f(a)) → a__f(g(f(a)))
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a__f(z0) → f(z0)
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mark(f(z0)) → a__f(z0)
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mark(a) → a
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mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
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A__F(f(a)) → c
S tuples:
A__F(f(a)) → c
K tuples:
MARK(f(z0)) → c2(A__F(z0))
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MARK(g(z0)) → c4(MARK(z0))
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK, A__F
Compound Symbols:
c2, c4, c
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(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
A__F(f(a)) → c
Now S is empty
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(10) BOUNDS(O(1), O(1))
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