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(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))

S tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

(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(c, c) → c4(F(a, a))

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

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))

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

POL(F(x₁, x₂)) = x₁   
POL(a) = 0   
POL(b) = [5]   
POL(c) = [2]   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(s(x₁)) = x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))

S tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))

K tuples:

F(c, c) → c4(F(a, a))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

(5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(a, b) → c2(F(s(a), c))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))

S tuples:

F(s(z0), c) → c3(F(z0, c))

K tuples:

F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

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

Use forward instantiation to replace F(s(z0), c) → c3(F(z0, c)) by

F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(c, c) → c4(F(a, a))
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))

S tuples:

F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))

K tuples:

F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c4, c3

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

Removed 4 trailing nodes:

F(s(c), c) → c3(F(c, c))
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(s(s(y0)), c) → c3(F(s(y0), c))

S tuples:

F(s(s(y0)), c) → c3(F(s(y0), c))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3

(11) 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(s(s(y0)), c) → c3(F(s(y0), c))

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

F(s(s(y0)), c) → c3(F(s(y0), c))

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

POL(F(x₁, x₂)) = x₁   
POL(c) = 0   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)

Tuples:

F(s(s(y0)), c) → c3(F(s(y0), c))

S tuples:none
K tuples:

F(s(s(y0)), c) → c3(F(s(y0), c))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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