YES(O(1), O(n^1)) 0.00/0.77 YES(O(1), O(n^1)) 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.78 0.00/0.78 0.00/0.78
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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(a, empty) → g(a, empty) 0.00/0.78
f(a, cons(x, k)) → f(cons(x, a), k) 0.00/0.78
g(empty, d) → d 0.00/0.78
g(cons(x, k), d) → g(k, cons(x, d))

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(z0, empty) → g(z0, empty) 0.00/0.78
f(z0, cons(z1, z2)) → f(cons(z1, z0), z2) 0.00/0.78
g(empty, z0) → z0 0.00/0.78
g(cons(z0, z1), z2) → g(z1, cons(z0, z2))
Tuples:

F(z0, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
S tuples:

F(z0, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, 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, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.78

POL(F(x1, x2)) = [2] + x1 + [2]x2    0.00/0.78
POL(G(x1, x2)) = [3] + x1    0.00/0.78
POL(c(x1)) = x1    0.00/0.78
POL(c1(x1)) = x1    0.00/0.78
POL(c3(x1)) = x1    0.00/0.78
POL(cons(x1, x2)) = [2] + x2    0.00/0.78
POL(empty) = [1]   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, empty) → g(z0, empty) 0.00/0.78
f(z0, cons(z1, z2)) → f(cons(z1, z0), z2) 0.00/0.78
g(empty, z0) → z0 0.00/0.78
g(cons(z0, z1), z2) → g(z1, cons(z0, z2))
Tuples:

F(z0, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
S tuples:none
K tuples:

F(z0, empty) → c(G(z0, empty)) 0.00/0.78
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2)) 0.00/0.78
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

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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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0.00/0.83 EOF