The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(z0))) → lastbit(z0)
conv(0) → cons(nil, 0)
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))

Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0)))

S tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0)))

K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

(3) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0))) by

CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0)))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(z0))) → lastbit(z0)
conv(0) → cons(nil, 0)
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))

Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0)))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

S tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0)))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

(5) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)

Removed 1 of 4 dangling nodes:

CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0)))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(z0))) → lastbit(z0)
conv(0) → cons(nil, 0)
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))

Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

S tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

(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.

CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

We considered the (Usable) Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))

And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(CONV(x₁)) = x₁   
POL(HALF(x₁)) = 0   
POL(LASTBIT(x₁)) = 0   
POL(c2(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(half(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(z0))) → lastbit(z0)
conv(0) → cons(nil, 0)
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))

Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

S tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))

K tuples:

CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))

We considered the (Usable) Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))

And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONV(x₁)) = x₁²   
POL(HALF(x₁)) = [2]x₁   
POL(LASTBIT(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(half(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(z0))) → lastbit(z0)
conv(0) → cons(nil, 0)
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))

Tuples:

HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))

S tuples:none
K tuples:

CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
HALF(s(s(z0))) → c2(HALF(z0))
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))

Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))