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

(0) Obligation:

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

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(not(not(z0)), z1, not(z2)) → and(z1, band(z0, z2), z0)

Tuples:

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

S tuples:

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

K tuples:none
Defined Rule Symbols:

and

Defined Pair Symbols:

AND

Compound Symbols:

c

(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.

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

We considered the (Usable) Rules:none
And the Tuples:

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

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

POL(AND(x₁, x₂, x₃)) = [4]x₁ + [5]x₂   
POL(band(x₁, x₂)) = [3]   
POL(c(x₁)) = x₁   
POL(not(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(not(not(z0)), z1, not(z2)) → and(z1, band(z0, z2), z0)

Tuples:

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

S tuples:none
K tuples:

AND(not(not(z0)), z1, not(z2)) → c(AND(z1, band(z0, z2), z0))

Defined Rule Symbols:

and

Defined Pair Symbols:

AND

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))