YES(O(1), O(n^2)) 0.00/0.87 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
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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0.00/0.90 0.00/0.90
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0.00/0.90

(0) Obligation:

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

norm(nil) → 0 0.00/0.90
norm(g(x, y)) → s(norm(x)) 0.00/0.90
f(x, nil) → g(nil, x) 0.00/0.90
f(x, g(y, z)) → g(f(x, y), z) 0.00/0.90
rem(nil, y) → nil 0.00/0.90
rem(g(x, y), 0) → g(x, y) 0.00/0.90
rem(g(x, y), s(z)) → rem(x, z)

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

Converted CpxTRS to CDT
0.00/0.90
0.00/0.90

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0 0.00/0.90
norm(g(z0, z1)) → s(norm(z0)) 0.00/0.90
f(z0, nil) → g(nil, z0) 0.00/0.90
f(z0, g(z1, z2)) → g(f(z0, z1), z2) 0.00/0.90
rem(nil, z0) → nil 0.00/0.90
rem(g(z0, z1), 0) → g(z0, z1) 0.00/0.90
rem(g(z0, z1), s(z2)) → rem(z0, z2)
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
S tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
K tuples:none
Defined Rule Symbols:

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

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0.00/0.90

(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.

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

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.90

POL(F(x1, x2)) = 0    0.00/0.90
POL(NORM(x1)) = x1    0.00/0.90
POL(REM(x1, x2)) = x2    0.00/0.90
POL(c1(x1)) = x1    0.00/0.90
POL(c3(x1)) = x1    0.00/0.90
POL(c6(x1)) = x1    0.00/0.90
POL(g(x1, x2)) = [1] + x1    0.00/0.90
POL(s(x1)) = x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0 0.00/0.90
norm(g(z0, z1)) → s(norm(z0)) 0.00/0.90
f(z0, nil) → g(nil, z0) 0.00/0.90
f(z0, g(z1, z2)) → g(f(z0, z1), z2) 0.00/0.90
rem(nil, z0) → nil 0.00/0.90
rem(g(z0, z1), 0) → g(z0, z1) 0.00/0.90
rem(g(z0, z1), s(z2)) → rem(z0, z2)
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
S tuples:

F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
K tuples:

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

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

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(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.

F(z0, g(z1, z2)) → c3(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.90

POL(F(x1, x2)) = [2]x2    0.00/0.90
POL(NORM(x1)) = [5]x1    0.00/0.90
POL(REM(x1, x2)) = 0    0.00/0.90
POL(c1(x1)) = x1    0.00/0.90
POL(c3(x1)) = x1    0.00/0.90
POL(c6(x1)) = x1    0.00/0.90
POL(g(x1, x2)) = [1] + x1    0.00/0.90
POL(s(x1)) = [1] + x1   
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0.00/0.90

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0 0.00/0.90
norm(g(z0, z1)) → s(norm(z0)) 0.00/0.90
f(z0, nil) → g(nil, z0) 0.00/0.90
f(z0, g(z1, z2)) → g(f(z0, z1), z2) 0.00/0.90
rem(nil, z0) → nil 0.00/0.90
rem(g(z0, z1), 0) → g(z0, z1) 0.00/0.90
rem(g(z0, z1), s(z2)) → rem(z0, z2)
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
S tuples:

REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
K tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1))
Defined Rule Symbols:

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

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(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.

REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.90

POL(F(x1, x2)) = 0    0.00/0.90
POL(NORM(x1)) = 0    0.00/0.90
POL(REM(x1, x2)) = x22    0.00/0.90
POL(c1(x1)) = x1    0.00/0.90
POL(c3(x1)) = x1    0.00/0.90
POL(c6(x1)) = x1    0.00/0.90
POL(g(x1, x2)) = 0    0.00/0.90
POL(s(x1)) = [1] + x1   
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0.00/0.90

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0 0.00/0.90
norm(g(z0, z1)) → s(norm(z0)) 0.00/0.90
f(z0, nil) → g(nil, z0) 0.00/0.90
f(z0, g(z1, z2)) → g(f(z0, z1), z2) 0.00/0.90
rem(nil, z0) → nil 0.00/0.90
rem(g(z0, z1), 0) → g(z0, z1) 0.00/0.90
rem(g(z0, z1), s(z2)) → rem(z0, z2)
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
S tuples:none
K tuples:

NORM(g(z0, z1)) → c1(NORM(z0)) 0.00/0.90
F(z0, g(z1, z2)) → c3(F(z0, z1)) 0.00/0.90
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))
Defined Rule Symbols:

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

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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