YES(O(1), O(n^2)) 0.00/1.00 YES(O(1), O(n^2)) 0.00/1.02 0.00/1.02 0.00/1.02 0.00/1.02 0.00/1.02 0.00/1.02 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/1.02 0.00/1.02 0.00/1.02
0.00/1.02
0.00/1.02
0.00/1.02
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
0.00/1.02 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/1.02 
↳2 CdtProblem
0.00/1.02 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/1.02 
↳4 CdtProblem
0.00/1.02 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/1.02 
↳6 CdtProblem
0.00/1.02 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
0.00/1.02 
↳8 CdtProblem
0.00/1.02 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/1.02 
↳10 BOUNDS(O(1), O(1))
0.00/1.02
0.00/1.02 0.00/1.02
0.00/1.02
0.00/1.02

(0) Obligation:

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

minus(X, 0) → X 0.00/1.02
minus(s(X), s(Y)) → p(minus(X, Y)) 0.00/1.02
p(s(X)) → X 0.00/1.02
div(0, s(Y)) → 0 0.00/1.02
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

Rewrite Strategy: INNERMOST
0.00/1.02
0.00/1.02

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

Converted CpxTRS to CDT
0.00/1.02
0.00/1.02

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0 0.00/1.02
div(0, s(z0)) → 0 0.00/1.02
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1)) 0.00/1.02
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1)) 0.00/1.02
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

MINUS, DIV

Compound Symbols:

c1, c4

0.00/1.02
0.00/1.02

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

Removed 1 trailing tuple parts
0.00/1.02
0.00/1.02

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0 0.00/1.02
div(0, s(z0)) → 0 0.00/1.02
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

0.00/1.02
0.00/1.02

(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.

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0
And the Tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.02

POL(0) = 0    0.00/1.02
POL(DIV(x1, x2)) = [2]x1    0.00/1.02
POL(MINUS(x1, x2)) = 0    0.00/1.02
POL(c1(x1)) = x1    0.00/1.02
POL(c4(x1, x2)) = x1 + x2    0.00/1.02
POL(minus(x1, x2)) = x1    0.00/1.02
POL(p(x1)) = x1    0.00/1.02
POL(s(x1)) = [1] + x1   
0.00/1.02
0.00/1.02

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0 0.00/1.02
div(0, s(z0)) → 0 0.00/1.02
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

0.00/1.02
0.00/1.02

(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.

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0
And the Tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.02

POL(0) = [3]    0.00/1.02
POL(DIV(x1, x2)) = x12    0.00/1.02
POL(MINUS(x1, x2)) = x1    0.00/1.02
POL(c1(x1)) = x1    0.00/1.02
POL(c4(x1, x2)) = x1 + x2    0.00/1.02
POL(minus(x1, x2)) = x1    0.00/1.02
POL(p(x1)) = x1    0.00/1.02
POL(s(x1)) = [2] + x1   
0.00/1.02
0.00/1.02

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 0.00/1.02
minus(s(z0), s(z1)) → p(minus(z0, z1)) 0.00/1.02
p(s(z0)) → z0 0.00/1.02
div(0, s(z0)) → 0 0.00/1.02
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:none
K tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.02
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

0.00/1.02
0.00/1.02

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

The set S is empty
0.00/1.02
0.00/1.02

(10) BOUNDS(O(1), O(1))

0.00/1.02
0.00/1.02
0.00/1.07 EOF