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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
odd(Cons(x, xs)) → even(xs)
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odd(Nil) → False
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even(Cons(x, xs)) → odd(xs)
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notEmpty(Cons(x, xs)) → True
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notEmpty(Nil) → False
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even(Nil) → True
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evenodd(x) → even(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:
odd(Cons(z0, z1)) → even(z1)
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odd(Nil) → False
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even(Cons(z0, z1)) → odd(z1)
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even(Nil) → True
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notEmpty(Cons(z0, z1)) → True
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notEmpty(Nil) → False
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evenodd(z0) → even(z0)
Tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
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EVENODD(z0) → c6(EVEN(z0))
S tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
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EVENODD(z0) → c6(EVEN(z0))
K tuples:none
Defined Rule Symbols:
odd, even, notEmpty, evenodd
Defined Pair Symbols:
ODD, EVEN, EVENODD
Compound Symbols:
c, c2, c6
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(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
EVENODD(z0) → c6(EVEN(z0))
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
odd(Cons(z0, z1)) → even(z1)
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odd(Nil) → False
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even(Cons(z0, z1)) → odd(z1)
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even(Nil) → True
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notEmpty(Cons(z0, z1)) → True
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notEmpty(Nil) → False
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evenodd(z0) → even(z0)
Tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
S tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
K tuples:none
Defined Rule Symbols:
odd, even, notEmpty, evenodd
Defined Pair Symbols:
ODD, EVEN
Compound Symbols:
c, 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.
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
We considered the (Usable) Rules:none
And the Tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(Cons(x1, x2)) = [4] + x2
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POL(EVEN(x1)) = [2] + [2]x1
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POL(ODD(x1)) = [4] + [2]x1
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POL(c(x1)) = x1
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POL(c2(x1)) = x1
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
odd(Cons(z0, z1)) → even(z1)
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odd(Nil) → False
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even(Cons(z0, z1)) → odd(z1)
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even(Nil) → True
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notEmpty(Cons(z0, z1)) → True
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notEmpty(Nil) → False
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evenodd(z0) → even(z0)
Tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
S tuples:none
K tuples:
ODD(Cons(z0, z1)) → c(EVEN(z1))
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EVEN(Cons(z0, z1)) → c2(ODD(z1))
Defined Rule Symbols:
odd, even, notEmpty, evenodd
Defined Pair Symbols:
ODD, EVEN
Compound Symbols:
c, c2
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(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
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(8) BOUNDS(O(1), O(1))
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