### (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 Cpx (relative) TRS 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) 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(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:none
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) CdtRhsSimplificationProcessorProof (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) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

### (11) 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(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)) = [4]x1 + x2
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [5] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

The set S is empty