### (0) Obligation:

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

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))
g(s(f(z0))) → g(f(z0))
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
G(s(f(z0))) → c3(G(f(z0)), F(z0))
S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
G(s(f(z0))) → c3(G(f(z0)), F(z0))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c3

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

G(s(f(z0))) → c3(G(f(z0)), F(z0))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))
g(s(f(z0))) → g(f(z0))
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

### (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.

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
We considered the (Usable) Rules:none
And the Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

### (9) 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(s(z0), z1)) → c1(F(c(z0, s(z1))))
We considered the (Usable) Rules:none
And the Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = [4]x1
POL(G(x1)) = 0
POL(c(x1, x2)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [4] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
S tuples:none
K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

The set S is empty