The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

(0) Obligation:

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

S tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

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.

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))

We considered the (Usable) Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

And the Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

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

POL(F(x₁, x₂)) = [4] + [2]x₂   
POL(c(x₁)) = [5] + x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(f(x₁, x₂)) = [3] + [5]x₁   
POL(s(x₁)) = 0   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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

Use instantiation to replace F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) by

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(s(z0), z1) → c2(F(z0, s(c(z1))))

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

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

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

POL(F(x₁, x₂)) = x₁ + x₁²   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

S tuples:none
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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