The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

(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:

G(z0, z0) → c5
F(a) → c1
F(c) → c2

(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(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

(12) BOUNDS(1, 1)