YES(O(1), O(n^1)) 0.00/0.74 YES(O(1), O(n^1)) 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.75 0.00/0.75 0.00/0.75
0.00/0.75 0.00/0.75 0.00/0.75
0.00/0.75
0.00/0.75

(0) Obligation:

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

f(x, x) → a 0.00/0.75
f(g(x), y) → f(x, y)

Rewrite Strategy: INNERMOST
0.00/0.75
0.00/0.75

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

Converted CpxTRS to CDT
0.00/0.75
0.00/0.75

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, z0) → a 0.00/0.75
f(g(z0), z1) → f(z0, z1)
Tuples:

F(g(z0), z1) → c1(F(z0, z1))
S tuples:

F(g(z0), z1) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

0.00/0.75
0.00/0.75

(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.

F(g(z0), z1) → c1(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(g(z0), z1) → c1(F(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.75

POL(F(x1, x2)) = [3]x1    0.00/0.75
POL(c1(x1)) = x1    0.00/0.75
POL(g(x1)) = [1] + x1   
0.00/0.75
0.00/0.75

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, z0) → a 0.00/0.75
f(g(z0), z1) → f(z0, z1)
Tuples:

F(g(z0), z1) → c1(F(z0, z1))
S tuples:none
K tuples:

F(g(z0), z1) → c1(F(z0, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

0.00/0.75
0.00/0.75

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

The set S is empty
0.00/0.75
0.00/0.75

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

0.00/0.75
0.00/0.75
0.00/0.77 EOF