YES(O(1), O(n^3)) 5.20/1.86 YES(O(1), O(n^3)) 5.20/1.89 5.20/1.89 5.20/1.89 5.20/1.89 5.20/1.89 5.20/1.89 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 5.20/1.89 5.20/1.89 5.20/1.89
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

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^2))))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))))
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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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5.20/1.89

(0) Obligation:

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

times(x, 0) → 0 5.20/1.89
times(x, s(y)) → plus(times(x, y), x) 5.20/1.89
plus(x, 0) → x 5.20/1.89
plus(0, x) → x 5.20/1.89
plus(x, s(y)) → s(plus(x, y)) 5.20/1.89
plus(s(x), y) → s(plus(x, y))

Rewrite Strategy: INNERMOST
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5.20/1.89

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

Converted CpxTRS to CDT
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5.20/1.89

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

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5.20/1.89

(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.

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
We considered the (Usable) Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.20/1.89

POL(0) = [3]    5.20/1.89
POL(PLUS(x1, x2)) = [1]    5.20/1.89
POL(TIMES(x1, x2)) = [2]x2    5.20/1.89
POL(c1(x1, x2)) = x1 + x2    5.20/1.89
POL(c4(x1)) = x1    5.20/1.89
POL(c5(x1)) = x1    5.20/1.89
POL(plus(x1, x2)) = [5]    5.20/1.89
POL(s(x1)) = [2] + x1    5.20/1.89
POL(times(x1, x2)) = 0   
5.20/1.89
5.20/1.89

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

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(5) 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.

PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
We considered the (Usable) Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.20/1.89

POL(0) = 0    5.20/1.89
POL(PLUS(x1, x2)) = [2]x2    5.20/1.89
POL(TIMES(x1, x2)) = x22 + x1·x2    5.20/1.89
POL(c1(x1, x2)) = x1 + x2    5.20/1.89
POL(c4(x1)) = x1    5.20/1.89
POL(c5(x1)) = x1    5.20/1.89
POL(plus(x1, x2)) = [3] + [3]x22 + [3]x1·x2    5.20/1.89
POL(s(x1)) = [2] + x1    5.20/1.89
POL(times(x1, x2)) = [3]x12   
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5.20/1.89

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

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5.20/1.89

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
We considered the (Usable) Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.20/1.89

POL(0) = 0    5.20/1.89
POL(PLUS(x1, x2)) = x1 + x22    5.20/1.89
POL(TIMES(x1, x2)) = x12·x2 + x1·x22    5.20/1.89
POL(c1(x1, x2)) = x1 + x2    5.20/1.89
POL(c4(x1)) = x1    5.20/1.89
POL(c5(x1)) = x1    5.20/1.89
POL(plus(x1, x2)) = x1 + x2    5.20/1.89
POL(s(x1)) = [1] + x1    5.20/1.89
POL(times(x1, x2)) = x1 + x1·x2   
5.20/1.89
5.20/1.89

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0 5.20/1.89
times(z0, s(z1)) → plus(times(z0, z1), z0) 5.20/1.89
plus(z0, 0) → z0 5.20/1.89
plus(0, z0) → z0 5.20/1.89
plus(z0, s(z1)) → s(plus(z0, z1)) 5.20/1.89
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:none
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1)) 5.20/1.89
PLUS(z0, s(z1)) → c4(PLUS(z0, z1)) 5.20/1.89
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

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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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5.49/1.94 EOF