0.00/0.80
0.00/0.80
(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0, y) → 0
0.00/0.80
f(s(x), y) → f(f(x, y), 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:
f(0, z0) → 0
0.00/0.80
f(s(z0), z1) → f(f(z0, z1), z1)
Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
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.
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
We considered the (Usable) Rules:
f(0, z0) → 0
0.00/0.80
f(s(z0), z1) → f(f(z0, z1), z1)
And the Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
0.00/0.80
POL(0) = 0
0.00/0.80
POL(F(x1, x2)) = [1] + [4]x1
0.00/0.80
POL(c1(x1, x2)) = x1 + x2
0.00/0.80
POL(f(x1, x2)) = 0
0.00/0.80
POL(s(x1)) = [4] + x1
0.00/0.80
0.00/0.80
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0, z0) → 0
0.00/0.80
f(s(z0), z1) → f(f(z0, z1), z1)
Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:none
K tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
0.00/0.80
0.00/0.80
(5) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
0.00/0.80
0.00/0.80
(6) BOUNDS(O(1), O(1))
0.00/0.80