### (0) Obligation:

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

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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(a) → f(c(a))
f(c(z0)) → z0
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(z0)) → z0
f(c(b)) → f(d(a))
e(g(z0)) → e(z0)
Tuples:

F(a) → c1(F(c(a)))
F(c(z0)) → c2
F(c(a)) → c3(F(d(b)))
F(a) → c4(F(d(a)))
F(d(z0)) → c5
F(c(b)) → c6(F(d(a)))
E(g(z0)) → c7(E(z0))
S tuples:

F(a) → c1(F(c(a)))
F(c(z0)) → c2
F(c(a)) → c3(F(d(b)))
F(a) → c4(F(d(a)))
F(d(z0)) → c5
F(c(b)) → c6(F(d(a)))
E(g(z0)) → c7(E(z0))
K tuples:none
Defined Rule Symbols:

f, e

Defined Pair Symbols:

F, E

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7

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

Removed 6 trailing nodes:

F(a) → c1(F(c(a)))
F(c(z0)) → c2
F(c(b)) → c6(F(d(a)))
F(c(a)) → c3(F(d(b)))
F(a) → c4(F(d(a)))
F(d(z0)) → c5

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → f(c(a))
f(c(z0)) → z0
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(z0)) → z0
f(c(b)) → f(d(a))
e(g(z0)) → e(z0)
Tuples:

E(g(z0)) → c7(E(z0))
S tuples:

E(g(z0)) → c7(E(z0))
K tuples:none
Defined Rule Symbols:

f, e

Defined Pair Symbols:

E

Compound Symbols:

c7

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(a) → f(c(a))
f(c(z0)) → z0
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(z0)) → z0
f(c(b)) → f(d(a))
e(g(z0)) → e(z0)

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

E(g(z0)) → c7(E(z0))
S tuples:

E(g(z0)) → c7(E(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

E

Compound Symbols:

c7

### (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.

E(g(z0)) → c7(E(z0))
We considered the (Usable) Rules:none
And the Tuples:

E(g(z0)) → c7(E(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(E(x1)) = [5]x1
POL(c7(x1)) = x1
POL(g(x1)) = [1] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

E(g(z0)) → c7(E(z0))
S tuples:none
K tuples:

E(g(z0)) → c7(E(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

E

Compound Symbols:

c7

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

The set S is empty