### (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 Cpx (relative) TRS 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) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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)

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

### (5) CdtRuleRemovalProof (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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

### (7) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) 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))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c3, c4

### (9) 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))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c1, c2, c4, c3

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

Removed 4 trailing nodes:

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

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c3

### (13) CdtRuleRemovalProof (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

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c3

### (15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty