YES(O(1), O(n^1)) 0.00/0.92 YES(O(1), O(n^1)) 0.00/0.94 0.00/0.94 0.00/0.94 0.00/0.94 0.00/0.94 0.00/0.94 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.94 0.00/0.94 0.00/0.94
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳4 CdtProblem
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↳5 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳6 BOUNDS(O(1), O(1))
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(0) Obligation:

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

f(0, y) → y 0.00/0.94
f(x, 0) → x 0.00/0.94
f(i(x), y) → i(x) 0.00/0.94
f(f(x, y), z) → f(x, f(y, z)) 0.00/0.94
f(g(x, y), z) → g(f(x, z), f(y, z)) 0.00/0.94
f(1, g(x, y)) → x 0.00/0.94
f(2, g(x, y)) → y

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:

f(0, z0) → z0 0.00/0.94
f(z0, 0) → z0 0.00/0.94
f(i(z0), z1) → i(z0) 0.00/0.94
f(f(z0, z1), z2) → f(z0, f(z1, z2)) 0.00/0.94
f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2)) 0.00/0.94
f(1, g(z0, z1)) → z0 0.00/0.94
f(2, g(z0, z1)) → z1
Tuples:

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
S tuples:

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

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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.

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
We considered the (Usable) Rules:

f(z0, 0) → z0 0.00/0.94
f(i(z0), z1) → i(z0) 0.00/0.94
f(f(z0, z1), z2) → f(z0, f(z1, z2)) 0.00/0.94
f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2)) 0.00/0.94
f(1, g(z0, z1)) → z0 0.00/0.94
f(2, g(z0, z1)) → z1 0.00/0.94
f(0, z0) → z0
And the Tuples:

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.94

POL(0) = 0    0.00/0.94
POL(1) = 0    0.00/0.94
POL(2) = 0    0.00/0.94
POL(F(x1, x2)) = [1] + [4]x1    0.00/0.94
POL(c3(x1, x2)) = x1 + x2    0.00/0.94
POL(c4(x1, x2)) = x1 + x2    0.00/0.94
POL(f(x1, x2)) = [4] + x1 + x2    0.00/0.94
POL(g(x1, x2)) = [4] + x1 + x2    0.00/0.94
POL(i(x1)) = x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, z0) → z0 0.00/0.94
f(z0, 0) → z0 0.00/0.94
f(i(z0), z1) → i(z0) 0.00/0.94
f(f(z0, z1), z2) → f(z0, f(z1, z2)) 0.00/0.94
f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2)) 0.00/0.94
f(1, g(z0, z1)) → z0 0.00/0.94
f(2, g(z0, z1)) → z1
Tuples:

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
S tuples:none
K tuples:

F(f(z0, z1), z2) → c3(F(z0, f(z1, z2)), F(z1, z2)) 0.00/0.94
F(g(z0, z1), z2) → c4(F(z0, z2), F(z1, z2))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

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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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0.00/0.96 EOF