YES(O(1), O(n^1)) 0.00/0.76 YES(O(1), O(n^1)) 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.77 0.00/0.77 0.00/0.77
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(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
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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:

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

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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.

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 : 0.00/0.77

POL(AND(x1, x2, x3)) = [4]x1 + [4]x2    0.00/0.77
POL(band(x1, x2)) = 0    0.00/0.77
POL(c(x1)) = x1    0.00/0.77
POL(not(x1)) = [4] + x1   
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(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

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(5) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty
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(6) BOUNDS(O(1), O(1))

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0.00/0.79 EOF