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, a), y) → f(y, f(x, y))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, a), z1) → f(z1, f(z0, z1))

Tuples:

F(f(z0, a), z1) → c(F(z1, f(z0, z1)), F(z0, z1))

S tuples:

F(f(z0, a), z1) → c(F(z1, f(z0, z1)), F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

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

Use forward instantiation to replace F(f(z0, a), z1) → c(F(z1, f(z0, z1)), F(z0, z1)) by

F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a)))
F(f(f(y0, a), a), z1) → c(F(z1, f(f(y0, a), z1)), F(f(y0, a), z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, a), z1) → f(z1, f(z0, z1))

Tuples:

F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a)))
F(f(f(y0, a), a), z1) → c(F(z1, f(f(y0, a), z1)), F(f(y0, a), z1))

S tuples:

F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a)))
F(f(f(y0, a), a), z1) → c(F(z1, f(f(y0, a), z1)), F(f(y0, a), z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

F(f(f(y0, a), a), z1) → c(F(z1, f(f(y0, a), z1)), F(f(y0, a), z1))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, a), z1) → f(z1, f(z0, z1))

Tuples:

F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a)))

S tuples:

F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

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

Use narrowing to replace F(f(z0, a), f(y0, a)) → c(F(f(y0, a), f(z0, f(y0, a))), F(z0, f(y0, a))) by

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, a), z1) → f(z1, f(z0, z1))

Tuples:

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

S tuples:

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(9) 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(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

We considered the (Usable) Rules:none
And the Tuples:

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

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

POL(F(x₁, x₂)) = [2]x₁   
POL(a) = [3]   
POL(c(x₁)) = x₁   
POL(f(x₁, x₂)) = [1] + [4]x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0, a), z1) → f(z1, f(z0, z1))

Tuples:

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

S tuples:none
K tuples:

F(f(x0, a), f(x1, a)) → c(F(x0, f(x1, a)))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))