(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(x, y) → g(x, y)
g(h(x), y) → h(f(x, y))
g(h(x), y) → h(g(x, y))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, z1) → g(z0, z1)
g(h(z0), z1) → h(f(z0, z1))
g(h(z0), z1) → h(g(z0, z1))
Tuples:
F(z0, z1) → c(G(z0, z1))
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
S tuples:
F(z0, z1) → c(G(z0, z1))
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c, c1, c2
(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.
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
F(z0, z1) → c(G(z0, z1))
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = [2] + [2]x1
POL(G(x1, x2)) = [2] + [2]x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(h(x1)) = [4] + x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, z1) → g(z0, z1)
g(h(z0), z1) → h(f(z0, z1))
g(h(z0), z1) → h(g(z0, z1))
Tuples:
F(z0, z1) → c(G(z0, z1))
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
S tuples:
F(z0, z1) → c(G(z0, z1))
K tuples:
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c, c1, c2
(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(z0, z1) → c(G(z0, z1))
G(h(z0), z1) → c1(F(z0, z1))
G(h(z0), z1) → c2(G(z0, z1))
Now S is empty
(6) BOUNDS(O(1), O(1))