YES(O(1), O(n^1)) 0.00/0.78 YES(O(1), O(n^1)) 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 0.00/0.80 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.80 0.00/0.80 0.00/0.80
0.00/0.80
0.00/0.80
0.00/0.80
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.80 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳2 CdtProblem
0.00/0.80 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.80 
↳4 CdtProblem
0.00/0.80 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.80 
↳6 CdtProblem
0.00/0.80 
↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.80 
↳8 BOUNDS(O(1), O(1))
0.00/0.80
0.00/0.80 0.00/0.80
0.00/0.80
0.00/0.80

(0) Obligation:

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

+(0, y) → y 0.00/0.80
+(s(x), y) → s(+(x, y)) 0.00/0.80
-(0, y) → 0 0.00/0.80
-(x, 0) → x 0.00/0.80
-(s(x), s(y)) → -(x, y)

Rewrite Strategy: INNERMOST
0.00/0.80
0.00/0.80

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

Converted CpxTRS to CDT
0.00/0.80
0.00/0.80

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0 0.00/0.80
+(s(z0), z1) → s(+(z0, z1)) 0.00/0.80
-(0, z0) → 0 0.00/0.80
-(z0, 0) → z0 0.00/0.80
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

0.00/0.80
0.00/0.80

(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.

+'(s(z0), z1) → c1(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.80

POL(+'(x1, x2)) = x1    0.00/0.80
POL(-'(x1, x2)) = 0    0.00/0.80
POL(c1(x1)) = x1    0.00/0.80
POL(c4(x1)) = x1    0.00/0.80
POL(s(x1)) = [1] + x1   
0.00/0.80
0.00/0.80

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0 0.00/0.80
+(s(z0), z1) → s(+(z0, z1)) 0.00/0.80
-(0, z0) → 0 0.00/0.80
-(z0, 0) → z0 0.00/0.80
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

0.00/0.80
0.00/0.80

(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)) → c4(-'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.80

POL(+'(x1, x2)) = [5]x1    0.00/0.80
POL(-'(x1, x2)) = x1    0.00/0.80
POL(c1(x1)) = x1    0.00/0.80
POL(c4(x1)) = x1    0.00/0.80
POL(s(x1)) = [1] + x1   
0.00/0.80
0.00/0.80

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0 0.00/0.80
+(s(z0), z1) → s(+(z0, z1)) 0.00/0.80
-(0, z0) → 0 0.00/0.80
-(z0, 0) → z0 0.00/0.80
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:none
K tuples:

+'(s(z0), z1) → c1(+'(z0, z1)) 0.00/0.80
-'(s(z0), s(z1)) → c4(-'(z0, z1))
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

0.00/0.80
0.00/0.80

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

The set S is empty
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0.00/0.80

(8) BOUNDS(O(1), O(1))

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0.00/0.80
0.00/0.83 EOF