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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(X) → if(X, c, n__f(true))
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if(true, X, Y) → X
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if(false, X, Y) → activate(Y)
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f(X) → n__f(X)
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activate(n__f(X)) → f(X)
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activate(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:
f(z0) → if(z0, c, n__f(true))
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f(z0) → n__f(z0)
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if(true, z0, z1) → z0
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if(false, z0, z1) → activate(z1)
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activate(n__f(z0)) → f(z0)
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activate(z0) → z0
Tuples:
F(z0) → c1(IF(z0, c, n__f(true)))
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IF(false, z0, z1) → c4(ACTIVATE(z1))
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ACTIVATE(n__f(z0)) → c5(F(z0))
S tuples:
F(z0) → c1(IF(z0, c, n__f(true)))
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IF(false, z0, z1) → c4(ACTIVATE(z1))
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ACTIVATE(n__f(z0)) → c5(F(z0))
K tuples:none
Defined Rule Symbols:
f, if, activate
Defined Pair Symbols:
F, IF, ACTIVATE
Compound Symbols:
c1, c4, c5
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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.
IF(false, z0, z1) → c4(ACTIVATE(z1))
We considered the (Usable) Rules:none
And the Tuples:
F(z0) → c1(IF(z0, c, n__f(true)))
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IF(false, z0, z1) → c4(ACTIVATE(z1))
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ACTIVATE(n__f(z0)) → c5(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(ACTIVATE(x1)) = x1
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POL(F(x1)) = x1
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POL(IF(x1, x2, x3)) = x1 + x2 + x3
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POL(c) = 0
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POL(c1(x1)) = x1
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POL(c4(x1)) = x1
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POL(c5(x1)) = x1
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POL(false) = [2]
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POL(n__f(x1)) = x1
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POL(true) = 0
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0) → if(z0, c, n__f(true))
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f(z0) → n__f(z0)
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if(true, z0, z1) → z0
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if(false, z0, z1) → activate(z1)
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activate(n__f(z0)) → f(z0)
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activate(z0) → z0
Tuples:
F(z0) → c1(IF(z0, c, n__f(true)))
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IF(false, z0, z1) → c4(ACTIVATE(z1))
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ACTIVATE(n__f(z0)) → c5(F(z0))
S tuples:
F(z0) → c1(IF(z0, c, n__f(true)))
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ACTIVATE(n__f(z0)) → c5(F(z0))
K tuples:
IF(false, z0, z1) → c4(ACTIVATE(z1))
Defined Rule Symbols:
f, if, activate
Defined Pair Symbols:
F, IF, ACTIVATE
Compound Symbols:
c1, c4, c5
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(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
ACTIVATE(n__f(z0)) → c5(F(z0))
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F(z0) → c1(IF(z0, c, n__f(true)))
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IF(false, z0, z1) → c4(ACTIVATE(z1))
Now S is empty
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(6) BOUNDS(O(1), O(1))
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