YES(O(1), O(n^2)) 0.00/0.99 YES(O(1), O(n^2)) 2.52/1.01 2.52/1.01 2.52/1.01 2.52/1.01 2.52/1.01 2.52/1.01 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.52/1.01 2.52/1.01 2.52/1.01
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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))
2.52/1.01 
↳2 CdtProblem
2.52/1.01 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
2.52/1.01 
↳4 CdtProblem
2.52/1.01 
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
2.52/1.01 
↳6 CdtProblem
2.52/1.01 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.52/1.01 
↳8 CdtProblem
2.52/1.01 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
2.52/1.01 
↳10 CdtProblem
2.52/1.01 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
2.52/1.01 
↳12 BOUNDS(O(1), O(1))
2.52/1.01
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2.52/1.01

(0) Obligation:

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

and(tt, X) → activate(X) 2.52/1.01
plus(N, 0) → N 2.52/1.01
plus(N, s(M)) → s(plus(N, M)) 2.52/1.01
x(N, 0) → 0 2.52/1.01
x(N, s(M)) → plus(x(N, M), N) 2.52/1.01
activate(X) → X

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

Converted CpxTRS to CDT
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2.52/1.01

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0) 2.52/1.01
plus(z0, 0) → z0 2.52/1.01
plus(z0, s(z1)) → s(plus(z0, z1)) 2.52/1.01
x(z0, 0) → 0 2.52/1.01
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.01
activate(z0) → z0
Tuples:

AND(tt, z0) → c(ACTIVATE(z0)) 2.52/1.01
PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.01
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:

AND(tt, z0) → c(ACTIVATE(z0)) 2.52/1.01
PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.01
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
K tuples:none
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

AND, PLUS, X

Compound Symbols:

c, c2, c4

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2.52/1.01

(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts
2.52/1.02
2.52/1.02

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1)) 2.52/1.02
x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1)) 2.52/1.02
AND(tt, z0) → c
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1)) 2.52/1.02
AND(tt, z0) → c
K tuples:none
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

PLUS, X, AND

Compound Symbols:

c2, c4, c

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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

AND(tt, z0) → c
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1)) 2.52/1.02
x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
K tuples:none
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

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(7) 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.

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

x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1))
And the Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.52/1.02

POL(0) = [3]    2.52/1.02
POL(PLUS(x1, x2)) = [1]    2.52/1.02
POL(X(x1, x2)) = [4]x2    2.52/1.02
POL(c2(x1)) = x1    2.52/1.02
POL(c4(x1, x2)) = x1 + x2    2.52/1.02
POL(plus(x1, x2)) = [3] + [5]x1    2.52/1.02
POL(s(x1)) = [4] + x1    2.52/1.02
POL(x(x1, x2)) = 0   
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2.52/1.02

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1)) 2.52/1.02
x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

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2.52/1.02

(9) 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)) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:

x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1))
And the Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.52/1.02

POL(0) = 0    2.52/1.02
POL(PLUS(x1, x2)) = x2    2.52/1.02
POL(X(x1, x2)) = [2]x1·x2    2.52/1.02
POL(c2(x1)) = x1    2.52/1.02
POL(c4(x1, x2)) = x1 + x2    2.52/1.02
POL(plus(x1, x2)) = [3] + [3]x2 + [3]x12    2.52/1.02
POL(s(x1)) = [1] + x1    2.52/1.02
POL(x(x1, x2)) = [3]x12   
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2.52/1.02

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0) 2.52/1.02
plus(z0, 0) → z0 2.52/1.02
plus(z0, s(z1)) → s(plus(z0, z1)) 2.52/1.02
x(z0, 0) → 0 2.52/1.02
x(z0, s(z1)) → plus(x(z0, z1), z0) 2.52/1.02
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1)) 2.52/1.02
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:none
K tuples:

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1)) 2.52/1.02
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

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

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

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2.52/1.06 EOF