YES(O(1), O(n^2)) 0.00/0.89 YES(O(1), O(n^2)) 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 0.00/0.90 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.90 0.00/0.90 0.00/0.90
0.00/0.90 0.00/0.90 0.00/0.90
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(0) Obligation:

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

g(S(x), y) → g(x, S(y)) 0.00/0.90
f(y, S(x)) → f(S(y), x) 0.00/0.90
g(0, x2) → x2 0.00/0.90
f(x1, 0) → g(x1, 0)

Rewrite Strategy: INNERMOST
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0.00/0.90

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

Converted CpxTRS to CDT
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0.00/0.90

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1)) 0.00/0.90
g(0, z0) → z0 0.00/0.90
f(z0, S(z1)) → f(S(z0), z1) 0.00/0.90
f(z0, 0) → g(z0, 0)
Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.90
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.90
F(z0, 0) → c3(G(z0, 0))
S tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.90
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.90
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

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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(z0, 0) → c3(G(z0, 0))
We considered the (Usable) Rules:none
And the Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.90
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.91

POL(0) = [4]    0.00/0.91
POL(F(x1, x2)) = [3] + [4]x1 + [4]x2    0.00/0.91
POL(G(x1, x2)) = [3] + [2]x2    0.00/0.91
POL(S(x1)) = x1    0.00/0.91
POL(c(x1)) = x1    0.00/0.91
POL(c2(x1)) = x1    0.00/0.91
POL(c3(x1)) = x1   
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0.00/0.91

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1)) 0.00/0.91
g(0, z0) → z0 0.00/0.91
f(z0, S(z1)) → f(S(z0), z1) 0.00/0.91
f(z0, 0) → g(z0, 0)
Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
S tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1))
K tuples:

F(z0, 0) → c3(G(z0, 0))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

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0.00/0.91

(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(S(z0), z1) → c(G(z0, S(z1)))
We considered the (Usable) Rules:none
And the Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.91

POL(0) = [4]    0.00/0.91
POL(F(x1, x2)) = [1] + [4]x1 + [4]x2    0.00/0.91
POL(G(x1, x2)) = [5] + [4]x1 + [2]x2    0.00/0.91
POL(S(x1)) = [4] + x1    0.00/0.91
POL(c(x1)) = x1    0.00/0.91
POL(c2(x1)) = x1    0.00/0.91
POL(c3(x1)) = x1   
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0.00/0.91

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1)) 0.00/0.91
g(0, z0) → z0 0.00/0.91
f(z0, S(z1)) → f(S(z0), z1) 0.00/0.91
f(z0, 0) → g(z0, 0)
Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
S tuples:

F(z0, S(z1)) → c2(F(S(z0), z1))
K tuples:

F(z0, 0) → c3(G(z0, 0)) 0.00/0.91
G(S(z0), z1) → c(G(z0, S(z1)))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

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0.00/0.91

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(z0, S(z1)) → c2(F(S(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.91

POL(0) = 0    0.00/0.91
POL(F(x1, x2)) = [2]x22    0.00/0.91
POL(G(x1, x2)) = 0    0.00/0.91
POL(S(x1)) = [2] + x1    0.00/0.91
POL(c(x1)) = x1    0.00/0.91
POL(c2(x1)) = x1    0.00/0.91
POL(c3(x1)) = x1   
0.00/0.91
0.00/0.91

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1)) 0.00/0.91
g(0, z0) → z0 0.00/0.91
f(z0, S(z1)) → f(S(z0), z1) 0.00/0.91
f(z0, 0) → g(z0, 0)
Tuples:

G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1)) 0.00/0.91
F(z0, 0) → c3(G(z0, 0))
S tuples:none
K tuples:

F(z0, 0) → c3(G(z0, 0)) 0.00/0.91
G(S(z0), z1) → c(G(z0, S(z1))) 0.00/0.91
F(z0, S(z1)) → c2(F(S(z0), z1))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

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

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

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0.00/0.95 EOF