YES(O(1), O(n^3)) 11.99/3.88 YES(O(1), O(n^3)) 11.99/3.89 11.99/3.89 11.99/3.89 11.99/3.89 11.99/3.89 11.99/3.89 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 11.99/3.89 11.99/3.89 11.99/3.89
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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
11.99/3.89 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
11.99/3.89 
↳2 CdtProblem
11.99/3.89 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
11.99/3.89 
↳4 CdtProblem
11.99/3.89 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
11.99/3.89 
↳6 CdtProblem
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↳7 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
11.99/3.89 
↳8 CdtProblem
11.99/3.89 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))))
11.99/3.89 
↳10 CdtProblem
11.99/3.89 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳12 BOUNDS(O(1), O(1))
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11.99/3.89

(0) Obligation:

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

le(0, Y) → true 11.99/3.89
le(s(X), 0) → false 11.99/3.89
le(s(X), s(Y)) → le(X, Y) 11.99/3.89
minus(0, Y) → 0 11.99/3.89
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y) 11.99/3.89
ifMinus(true, s(X), Y) → 0 11.99/3.89
ifMinus(false, s(X), Y) → s(minus(X, Y)) 11.99/3.89
quot(0, s(Y)) → 0 11.99/3.89
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Rewrite Strategy: INNERMOST
11.99/3.89
11.99/3.89

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

Converted CpxTRS to CDT
11.99/3.89
11.99/3.89

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 11.99/3.89
le(s(z0), 0) → false 11.99/3.89
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.89
minus(0, z0) → 0 11.99/3.89
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.89
ifMinus(true, s(z0), z1) → 0 11.99/3.89
ifMinus(false, s(z0), z1) → s(minus(z0, z1)) 11.99/3.89
quot(0, s(z0)) → 0 11.99/3.89
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.89
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.89
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

11.99/3.89
11.99/3.89

(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.

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(0, z0) → 0 11.99/3.89
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.89
le(s(z0), 0) → false 11.99/3.89
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.89
le(0, z0) → true 11.99/3.89
ifMinus(true, s(z0), z1) → 0 11.99/3.89
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.89
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 11.99/3.89

POL(0) = 0    11.99/3.89
POL(IFMINUS(x1, x2, x3)) = 0    11.99/3.89
POL(LE(x1, x2)) = 0    11.99/3.89
POL(MINUS(x1, x2)) = 0    11.99/3.89
POL(QUOT(x1, x2)) = x1    11.99/3.89
POL(c2(x1)) = x1    11.99/3.89
POL(c4(x1, x2)) = x1 + x2    11.99/3.89
POL(c6(x1)) = x1    11.99/3.89
POL(c8(x1, x2)) = x1 + x2    11.99/3.89
POL(false) = 0    11.99/3.89
POL(ifMinus(x1, x2, x3)) = x2    11.99/3.89
POL(le(x1, x2)) = 0    11.99/3.89
POL(minus(x1, x2)) = x1    11.99/3.89
POL(s(x1)) = [1] + x1    11.99/3.89
POL(true) = 0   
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11.99/3.89

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 11.99/3.89
le(s(z0), 0) → false 11.99/3.89
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.89
minus(0, z0) → 0 11.99/3.89
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.89
ifMinus(true, s(z0), z1) → 0 11.99/3.89
ifMinus(false, s(z0), z1) → s(minus(z0, z1)) 11.99/3.89
quot(0, s(z0)) → 0 11.99/3.89
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.89
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

11.99/3.89
11.99/3.89

(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.

IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(0, z0) → 0 11.99/3.89
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.89
le(s(z0), 0) → false 11.99/3.89
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.89
le(0, z0) → true 11.99/3.89
ifMinus(true, s(z0), z1) → 0 11.99/3.89
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.89
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.89
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.89
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 11.99/3.89

POL(0) = 0    11.99/3.89
POL(IFMINUS(x1, x2, x3)) = x2    11.99/3.89
POL(LE(x1, x2)) = 0    11.99/3.89
POL(MINUS(x1, x2)) = x1    11.99/3.89
POL(QUOT(x1, x2)) = x12    11.99/3.89
POL(c2(x1)) = x1    11.99/3.89
POL(c4(x1, x2)) = x1 + x2    11.99/3.89
POL(c6(x1)) = x1    11.99/3.89
POL(c8(x1, x2)) = x1 + x2    11.99/3.89
POL(false) = 0    11.99/3.89
POL(ifMinus(x1, x2, x3)) = x2    11.99/3.89
POL(le(x1, x2)) = 0    11.99/3.89
POL(minus(x1, x2)) = x1    11.99/3.89
POL(s(x1)) = [2] + x1    11.99/3.89
POL(true) = 0   
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11.99/3.89

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 11.99/3.89
le(s(z0), 0) → false 11.99/3.89
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.89
minus(0, z0) → 0 11.99/3.89
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.89
ifMinus(true, s(z0), z1) → 0 11.99/3.89
ifMinus(false, s(z0), z1) → s(minus(z0, z1)) 11.99/3.89
quot(0, s(z0)) → 0 11.99/3.89
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

11.99/3.90
11.99/3.90

(7) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
11.99/3.90
11.99/3.90

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 11.99/3.90
le(s(z0), 0) → false 11.99/3.90
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.90
minus(0, z0) → 0 11.99/3.90
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.90
ifMinus(true, s(z0), z1) → 0 11.99/3.90
ifMinus(false, s(z0), z1) → s(minus(z0, z1)) 11.99/3.90
quot(0, s(z0)) → 0 11.99/3.90
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

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

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

11.99/3.90
11.99/3.90

(9) 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.

LE(s(z0), s(z1)) → c2(LE(z0, z1))
We considered the (Usable) Rules:

minus(0, z0) → 0 11.99/3.90
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.90
le(s(z0), 0) → false 11.99/3.90
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.90
le(0, z0) → true 11.99/3.90
ifMinus(true, s(z0), z1) → 0 11.99/3.90
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 11.99/3.90

POL(0) = 0    11.99/3.90
POL(IFMINUS(x1, x2, x3)) = x22    11.99/3.90
POL(LE(x1, x2)) = [1] + x1    11.99/3.90
POL(MINUS(x1, x2)) = [1] + x1 + x12    11.99/3.90
POL(QUOT(x1, x2)) = x13    11.99/3.90
POL(c2(x1)) = x1    11.99/3.90
POL(c4(x1, x2)) = x1 + x2    11.99/3.90
POL(c6(x1)) = x1    11.99/3.90
POL(c8(x1, x2)) = x1 + x2    11.99/3.90
POL(false) = 0    11.99/3.90
POL(ifMinus(x1, x2, x3)) = x2    11.99/3.90
POL(le(x1, x2)) = 0    11.99/3.90
POL(minus(x1, x2)) = x1    11.99/3.90
POL(s(x1)) = [1] + x1    11.99/3.90
POL(true) = 0   
11.99/3.90
11.99/3.90

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 11.99/3.90
le(s(z0), 0) → false 11.99/3.90
le(s(z0), s(z1)) → le(z0, z1) 11.99/3.90
minus(0, z0) → 0 11.99/3.90
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1) 11.99/3.90
ifMinus(true, s(z0), z1) → 0 11.99/3.90
ifMinus(false, s(z0), z1) → s(minus(z0, z1)) 11.99/3.90
quot(0, s(z0)) → 0 11.99/3.90
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 11.99/3.90
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1)) 11.99/3.90
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1)) 11.99/3.90
LE(s(z0), s(z1)) → c2(LE(z0, z1))
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

11.99/3.90
11.99/3.90

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

The set S is empty
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11.99/3.90

(12) BOUNDS(O(1), O(1))

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11.99/3.90
12.18/3.97 EOF