YES(O(1), O(n^1)) 0.00/0.72 YES(O(1), O(n^1)) 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.73 0.00/0.73 0.00/0.73
0.00/0.73
0.00/0.73
0.00/0.73
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.73 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.73 
↳2 CdtProblem
0.00/0.73 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.73 
↳4 CdtProblem
0.00/0.73 
↳5 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.73 
↳6 BOUNDS(O(1), O(1))
0.00/0.73
0.00/0.73 0.00/0.73
0.00/0.73
0.00/0.73

(0) Obligation:

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

minus(minus(x)) → x 0.00/0.73
minus(h(x)) → h(minus(x)) 0.00/0.73
minus(f(x, y)) → f(minus(y), minus(x))

Rewrite Strategy: INNERMOST
0.00/0.73
0.00/0.73

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

Converted CpxTRS to CDT
0.00/0.73
0.00/0.73

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0 0.00/0.73
minus(h(z0)) → h(minus(z0)) 0.00/0.73
minus(f(z0, z1)) → f(minus(z1), minus(z0))
Tuples:

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

0.00/0.73
0.00/0.73

(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.

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
We considered the (Usable) Rules:none
And the Tuples:

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.73

POL(MINUS(x1)) = [2]x1    0.00/0.73
POL(c1(x1)) = x1    0.00/0.73
POL(c2(x1, x2)) = x1 + x2    0.00/0.73
POL(f(x1, x2)) = [3] + x1 + x2    0.00/0.73
POL(h(x1)) = [1] + x1   
0.00/0.73
0.00/0.73

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0 0.00/0.73
minus(h(z0)) → h(minus(z0)) 0.00/0.73
minus(f(z0, z1)) → f(minus(z1), minus(z0))
Tuples:

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:none
K tuples:

MINUS(h(z0)) → c1(MINUS(z0)) 0.00/0.73
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

0.00/0.73
0.00/0.73

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

The set S is empty
0.00/0.73
0.00/0.73

(6) BOUNDS(O(1), O(1))

0.00/0.73
0.00/0.73
0.00/0.73 EOF