The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(0) Obligation:

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

f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))
F(g(f(z0))) → c1(F(g(z0)))

S tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))
F(g(f(z0))) → c1(F(g(z0)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(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(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))

We considered the (Usable) Rules:

f(g(f(z0))) → f(g(z0))

And the Tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))
F(g(f(z0))) → c1(F(g(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x₁)) = [4]x₁   
POL(c(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c1(x₁)) = x₁   
POL(f(x₁)) = [2] + [4]x₁   
POL(g(x₁)) = 0   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))
F(g(f(z0))) → c1(F(g(z0)))

S tuples:

F(g(f(z0))) → c1(F(g(z0)))

K tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0)) by

F(f(z0)) → c(F(g(f(g(z0)))), F(g(f(z0))), F(z0))
F(f(x0)) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(g(f(z0))) → c1(F(g(z0)))
F(f(z0)) → c(F(g(f(g(z0)))), F(g(f(z0))), F(z0))
F(f(x0)) → c

S tuples:

F(g(f(z0))) → c1(F(g(z0)))

K tuples:

F(f(z0)) → c(F(g(f(g(f(z0))))), F(g(f(z0))), F(z0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c, c

(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

F(f(x0)) → c

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(g(f(z0))) → c1(F(g(z0)))
F(f(z0)) → c(F(g(f(g(z0)))), F(g(f(z0))), F(z0))

S tuples:

F(g(f(z0))) → c1(F(g(z0)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c

(9) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)

Use forward instantiation to replace F(g(f(z0))) → c1(F(g(z0))) by

F(g(f(f(y0)))) → c1(F(g(f(y0))))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(f(z0)) → c(F(g(f(g(z0)))), F(g(f(z0))), F(z0))
F(g(f(f(y0)))) → c1(F(g(f(y0))))

S tuples:

F(g(f(f(y0)))) → c1(F(g(f(y0))))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

F(g(f(f(y0)))) → c1(F(g(f(y0))))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(g(f(g(f(z0)))))
f(g(f(z0))) → f(g(z0))

Tuples:

F(f(z0)) → c(F(g(f(g(z0)))), F(g(f(z0))), F(z0))

S tuples:none
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(14) BOUNDS(O(1), O(1))