YES(O(1), O(n^1)) 0.00/0.82 YES(O(1), O(n^1)) 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.84 0.00/0.84 0.00/0.84
0.00/0.84 0.00/0.84 0.00/0.84
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0.00/0.84

(0) Obligation:

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

g(f(x, y), z) → f(x, g(y, z)) 0.00/0.84
g(h(x, y), z) → g(x, f(y, z)) 0.00/0.84
g(x, h(y, z)) → h(g(x, y), z)

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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0.00/0.84

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(f(z0, z1), z2) → f(z0, g(z1, z2)) 0.00/0.84
g(h(z0, z1), z2) → g(z0, f(z1, z2)) 0.00/0.84
g(z0, h(z1, z2)) → h(g(z0, z1), z2)
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

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0.00/0.84

(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.

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.84

POL(G(x1, x2)) = [4]x1    0.00/0.84
POL(c(x1)) = x1    0.00/0.84
POL(c1(x1)) = x1    0.00/0.84
POL(c2(x1)) = x1    0.00/0.84
POL(f(x1, x2)) = [1] + x1 + x2    0.00/0.84
POL(h(x1, x2)) = [4] + x1   
0.00/0.84
0.00/0.84

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(f(z0, z1), z2) → f(z0, g(z1, z2)) 0.00/0.84
g(h(z0, z1), z2) → g(z0, f(z1, z2)) 0.00/0.84
g(z0, h(z1, z2)) → h(g(z0, z1), z2)
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(z0, h(z1, z2)) → c2(G(z0, z1))
K tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

0.00/0.84
0.00/0.84

(5) 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.

G(z0, h(z1, z2)) → c2(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.84

POL(G(x1, x2)) = [4]x1 + x2    0.00/0.84
POL(c(x1)) = x1    0.00/0.84
POL(c1(x1)) = x1    0.00/0.84
POL(c2(x1)) = x1    0.00/0.84
POL(f(x1, x2)) = [2] + x1 + x2    0.00/0.84
POL(h(x1, x2)) = [3] + x1 + x2   
0.00/0.84
0.00/0.84

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(f(z0, z1), z2) → f(z0, g(z1, z2)) 0.00/0.84
g(h(z0, z1), z2) → g(z0, f(z1, z2)) 0.00/0.84
g(z0, h(z1, z2)) → h(g(z0, z1), z2)
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:none
K tuples:

G(f(z0, z1), z2) → c(G(z1, z2)) 0.00/0.84
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2))) 0.00/0.84
G(z0, h(z1, z2)) → c2(G(z0, z1))
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

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

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

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0.00/0.88 EOF