YES(O(1), O(n^1)) 0.00/0.75 YES(O(1), O(n^1)) 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.77 0.00/0.77 0.00/0.77
0.00/0.77
0.00/0.77
0.00/0.77
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.77 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.77 
↳2 CdtProblem
0.00/0.77 
↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication)
0.00/0.77 
↳4 CdtProblem
0.00/0.77 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.77 
↳6 CdtProblem
0.00/0.77 
↳7 CdtKnowledgeProof (⇔)
0.00/0.77 
↳8 BOUNDS(O(1), O(1))
0.00/0.77
0.00/0.77 0.00/0.77
0.00/0.77
0.00/0.77

(0) Obligation:

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

g(0, f(x, x)) → x 0.00/0.77
g(x, s(y)) → g(f(x, y), 0) 0.00/0.77
g(s(x), y) → g(f(x, y), 0) 0.00/0.77
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

Rewrite Strategy: INNERMOST
0.00/0.77
0.00/0.77

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

Converted CpxTRS to CDT
0.00/0.77
0.00/0.77

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0, f(z0, z0)) → z0 0.00/0.77
g(z0, s(z1)) → g(f(z0, z1), 0) 0.00/0.77
g(s(z0), z1) → g(f(z0, z1), 0) 0.00/0.77
g(f(z0, z1), 0) → f(g(z0, 0), g(z1, 0))
Tuples:

G(z0, s(z1)) → c1(G(f(z0, z1), 0)) 0.00/0.77
G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
S tuples:

G(z0, s(z1)) → c1(G(f(z0, z1), 0)) 0.00/0.77
G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c1, c2, c3

0.00/0.77
0.00/0.77

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(z0, s(z1)) → c1(G(f(z0, z1), 0))
0.00/0.77
0.00/0.77

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0, f(z0, z0)) → z0 0.00/0.77
g(z0, s(z1)) → g(f(z0, z1), 0) 0.00/0.77
g(s(z0), z1) → g(f(z0, z1), 0) 0.00/0.77
g(f(z0, z1), 0) → f(g(z0, 0), g(z1, 0))
Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
S tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c2, c3

0.00/0.77
0.00/0.77

(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.

G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.77

POL(0) = 0    0.00/0.77
POL(G(x1, x2)) = [3] + [4]x1 + [4]x2    0.00/0.77
POL(c2(x1)) = x1    0.00/0.77
POL(c3(x1, x2)) = x1 + x2    0.00/0.77
POL(f(x1, x2)) = [4] + x1 + x2    0.00/0.77
POL(s(x1)) = [4] + x1   
0.00/0.77
0.00/0.77

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0, f(z0, z0)) → z0 0.00/0.77
g(z0, s(z1)) → g(f(z0, z1), 0) 0.00/0.77
g(s(z0), z1) → g(f(z0, z1), 0) 0.00/0.77
g(f(z0, z1), 0) → f(g(z0, 0), g(z1, 0))
Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
S tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
K tuples:

G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c2, c3

0.00/0.77
0.00/0.77

(7) CdtKnowledgeProof (EQUIVALENT transformation)

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

G(s(z0), z1) → c2(G(f(z0, z1), 0)) 0.00/0.77
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))
Now S is empty
0.00/0.77
0.00/0.77

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

0.00/0.77
0.00/0.77
0.00/0.79 EOF