### (0) Obligation:

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

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(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) → b
f(c) → d
f(g(z0, z1)) → g(f(z0), f(z1))
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(a) → c1
F(c) → c2
F(g(z0, z1)) → c3(G(f(z0), f(z1)), F(z0), F(z1))
F(h(z0, z1)) → c4(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c5
S tuples:

F(a) → c1
F(c) → c2
F(g(z0, z1)) → c3(G(f(z0), f(z1)), F(z0), F(z1))
F(h(z0, z1)) → c4(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c5
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

F(a) → c1
F(c) → c2
G(z0, z0) → c5

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → b
f(c) → d
f(g(z0, z1)) → g(f(z0), f(z1))
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(g(z0, z1)) → c3(G(f(z0), f(z1)), F(z0), F(z1))
F(h(z0, z1)) → c4(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
S tuples:

F(g(z0, z1)) → c3(G(f(z0), f(z1)), F(z0), F(z1))
F(h(z0, z1)) → c4(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → b
f(c) → d
f(g(z0, z1)) → g(f(z0), f(z1))
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
S tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(a) → b
f(c) → d
f(g(z0, z1)) → g(f(z0), f(z1))
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
S tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

### (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(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
We considered the (Usable) Rules:none
And the Tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = [2] + [4]x1
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2
POL(g(x1, x2)) = [4] + x1 + x2
POL(h(x1, x2)) = [4] + x1 + x2

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
S tuples:none
K tuples:

F(g(z0, z1)) → c3(F(z0), F(z1))
F(h(z0, z1)) → c4(F(z0), F(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

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

The set S is empty