YES(O(1), O(n^2)) 3.19/1.21 YES(O(1), O(n^2)) 3.19/1.23 3.19/1.23 3.19/1.23 3.19/1.23 3.19/1.23 3.19/1.23 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.19/1.23 3.19/1.23 3.19/1.23
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3.19/1.23
3.19/1.23
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
3.19/1.23 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
3.19/1.23 
↳2 CdtProblem
3.19/1.23 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.19/1.23 
↳4 CdtProblem
3.19/1.23 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.19/1.23 
↳6 CdtProblem
3.19/1.23 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
3.19/1.23 
↳8 CdtProblem
3.19/1.23 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
3.19/1.23 
↳10 BOUNDS(O(1), O(1))
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3.19/1.23 3.19/1.23
3.19/1.23
3.19/1.23

(0) Obligation:

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

exp(x, 0) → s(0) 3.19/1.23
exp(x, s(y)) → *(x, exp(x, y)) 3.19/1.23
*(0, y) → 0 3.19/1.23
*(s(x), y) → +(y, *(x, y)) 3.19/1.23
-(0, y) → 0 3.19/1.23
-(x, 0) → x 3.19/1.23
-(s(x), s(y)) → -(x, y)

Rewrite Strategy: INNERMOST
3.19/1.23
3.19/1.23

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

Converted CpxTRS to CDT
3.19/1.23
3.19/1.23

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0) 3.19/1.23
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.23
*(0, z0) → 0 3.19/1.23
*(s(z0), z1) → +(z1, *(z0, z1)) 3.19/1.23
-(0, z0) → 0 3.19/1.23
-(z0, 0) → z0 3.19/1.23
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.23
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.23
-'(s(z0), s(z1)) → c6(-'(z0, z1))
S tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.23
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.23
-'(s(z0), s(z1)) → c6(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

3.19/1.23
3.19/1.23

(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.

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
We considered the (Usable) Rules:

exp(z0, 0) → s(0) 3.19/1.23
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.23
*(0, z0) → 0 3.19/1.23
*(s(z0), z1) → +(z1, *(z0, z1))
And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.23
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.23
-'(s(z0), s(z1)) → c6(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.19/1.23

POL(*(x1, x2)) = [2] + [4]x1    3.19/1.23
POL(*'(x1, x2)) = [5]    3.19/1.23
POL(+(x1, x2)) = [4]    3.19/1.23
POL(-'(x1, x2)) = 0    3.19/1.23
POL(0) = 0    3.19/1.23
POL(EXP(x1, x2)) = [4]x2    3.19/1.23
POL(c1(x1, x2)) = x1 + x2    3.19/1.23
POL(c3(x1)) = x1    3.19/1.23
POL(c6(x1)) = x1    3.19/1.23
POL(exp(x1, x2)) = [4] + [4]x1 + [2]x2    3.19/1.23
POL(s(x1)) = [4] + x1   
3.19/1.23
3.19/1.23

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0) 3.19/1.23
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.23
*(0, z0) → 0 3.19/1.23
*(s(z0), z1) → +(z1, *(z0, z1)) 3.19/1.23
-(0, z0) → 0 3.19/1.23
-(z0, 0) → z0 3.19/1.23
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.23
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.23
-'(s(z0), s(z1)) → c6(-'(z0, z1))
S tuples:

*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.23
-'(s(z0), s(z1)) → c6(-'(z0, z1))
K tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

3.19/1.24
3.19/1.24

(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.

-'(s(z0), s(z1)) → c6(-'(z0, z1))
We considered the (Usable) Rules:

exp(z0, 0) → s(0) 3.19/1.24
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.24
*(0, z0) → 0 3.19/1.24
*(s(z0), z1) → +(z1, *(z0, z1))
And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.19/1.24

POL(*(x1, x2)) = 0    3.19/1.24
POL(*'(x1, x2)) = [1]    3.19/1.24
POL(+(x1, x2)) = 0    3.19/1.24
POL(-'(x1, x2)) = x1    3.19/1.24
POL(0) = 0    3.19/1.24
POL(EXP(x1, x2)) = [2]x2    3.19/1.24
POL(c1(x1, x2)) = x1 + x2    3.19/1.24
POL(c3(x1)) = x1    3.19/1.24
POL(c6(x1)) = x1    3.19/1.24
POL(exp(x1, x2)) = [4] + [2]x2    3.19/1.24
POL(s(x1)) = [2] + x1   
3.19/1.24
3.19/1.24

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0) 3.19/1.24
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.24
*(0, z0) → 0 3.19/1.24
*(s(z0), z1) → +(z1, *(z0, z1)) 3.19/1.24
-(0, z0) → 0 3.19/1.24
-(z0, 0) → z0 3.19/1.24
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1))
S tuples:

*'(s(z0), z1) → c3(*'(z0, z1))
K tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1))
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

3.19/1.24
3.19/1.24

(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.

*'(s(z0), z1) → c3(*'(z0, z1))
We considered the (Usable) Rules:

exp(z0, 0) → s(0) 3.19/1.24
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.24
*(0, z0) → 0 3.19/1.24
*(s(z0), z1) → +(z1, *(z0, z1))
And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.19/1.24

POL(*(x1, x2)) = [1] + [3]x1    3.19/1.24
POL(*'(x1, x2)) = [2]x1    3.19/1.24
POL(+(x1, x2)) = [3] + x2    3.19/1.24
POL(-'(x1, x2)) = x1 + x2    3.19/1.24
POL(0) = [1]    3.19/1.24
POL(EXP(x1, x2)) = x1·x2    3.19/1.24
POL(c1(x1, x2)) = x1 + x2    3.19/1.24
POL(c3(x1)) = x1    3.19/1.24
POL(c6(x1)) = x1    3.19/1.24
POL(exp(x1, x2)) = [3] + [3]x1 + x2 + x22    3.19/1.24
POL(s(x1)) = [3] + x1   
3.19/1.24
3.19/1.24

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0) 3.19/1.24
exp(z0, s(z1)) → *(z0, exp(z0, z1)) 3.19/1.24
*(0, z0) → 0 3.19/1.24
*(s(z0), z1) → +(z1, *(z0, z1)) 3.19/1.24
-(0, z0) → 0 3.19/1.24
-(z0, 0) → z0 3.19/1.24
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
*'(s(z0), z1) → c3(*'(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1))
S tuples:none
K tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1)) 3.19/1.24
-'(s(z0), s(z1)) → c6(-'(z0, z1)) 3.19/1.24
*'(s(z0), z1) → c3(*'(z0, z1))
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

3.19/1.24
3.19/1.24

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

The set S is empty
3.19/1.24
3.19/1.24

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

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3.19/1.24
3.49/1.31 EOF