0.00/0.72
0.00/0.72
(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
h(f(x), y) → f(g(x, y))
0.00/0.72
g(x, y) → h(x, y)
Rewrite Strategy: INNERMOST
0.00/0.72
0.00/0.72
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
0.00/0.72
0.00/0.72
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
h(f(z0), z1) → f(g(z0, z1))
0.00/0.72
g(z0, z1) → h(z0, z1)
Tuples:
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
S tuples:
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
K tuples:none
Defined Rule Symbols:
h, g
Defined Pair Symbols:
H, G
Compound Symbols:
c, c1
0.00/0.72
0.00/0.72
(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.
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
0.00/0.72
POL(G(x1, x2)) = [3] + [2]x1
0.00/0.72
POL(H(x1, x2)) = [2] + [2]x1
0.00/0.72
POL(c(x1)) = x1
0.00/0.72
POL(c1(x1)) = x1
0.00/0.72
POL(f(x1)) = [4] + x1
0.00/0.72
0.00/0.72
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
h(f(z0), z1) → f(g(z0, z1))
0.00/0.72
g(z0, z1) → h(z0, z1)
Tuples:
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
S tuples:none
K tuples:
H(f(z0), z1) → c(G(z0, z1))
0.00/0.72
G(z0, z1) → c1(H(z0, z1))
Defined Rule Symbols:
h, g
Defined Pair Symbols:
H, G
Compound Symbols:
c, c1
0.00/0.72
0.00/0.72
(5) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
0.00/0.72
0.00/0.72
(6) BOUNDS(O(1), O(1))
0.00/0.72