(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(f(x)) → g(f(x))
g(g(x)) → f(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(f(z0)) → g(f(z0))
g(g(z0)) → f(z0)
Tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
G(g(z0)) → c1(F(z0))
S tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
G(g(z0)) → c1(F(z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c, c1
(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
G(g(z0)) → c1(F(z0))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(f(z0)) → g(f(z0))
g(g(z0)) → f(z0)
Tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
S tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F
Compound Symbols:
c
(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(f(z0)) → c(G(f(z0)), F(z0))
We considered the (Usable) Rules:none
And the Tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = [2]x12
POL(G(x1)) = 0
POL(c(x1, x2)) = x1 + x2
POL(f(x1)) = [2] + [3]x12
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(f(z0)) → g(f(z0))
g(g(z0)) → f(z0)
Tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
S tuples:none
K tuples:
F(f(z0)) → c(G(f(z0)), F(z0))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F
Compound Symbols:
c
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))