### (0) Obligation:

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

f(S(x), x2) → f(x2, x)
f(0, x2) → 0

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0
Tuples:

F(S(z0), z1) → c(F(z1, z0))
F(0, z0) → c1
S tuples:

F(S(z0), z1) → c(F(z1, z0))
F(0, z0) → c1
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(0, z0) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:

F(S(z0), z1) → c(F(z1, z0))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:

F(S(z0), z1) → c(F(z1, z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

### (7) 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) → c(F(z1, z0))
We considered the (Usable) Rules:none
And the Tuples:

F(S(z0), z1) → c(F(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = [2]x1 + [2]x2
POL(S(x1)) = [1] + x1
POL(c(x1)) = x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:none
K tuples:

F(S(z0), z1) → c(F(z1, z0))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty