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

(0) Obligation:

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

b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
c(x1) → g(x1)
d(d(x1)) → c(f(x1))
d(d(d(x1))) → g(c(x1))
f(x1) → a(g(x1))
g(x1) → d(a(b(x1)))
g(g(x1)) → b(c(x1))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(c(z0)) → c2(D(d(d(z0))), D(d(z0)), D(z0))
C(z0) → c3(G(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
D(d(d(z0))) → c5(G(c(z0)), C(z0))
F(z0) → c6(G(z0))
G(z0) → c7(D(a(b(z0))), B(z0))
G(g(z0)) → c8(B(c(z0)), C(z0))

S tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(c(z0)) → c2(D(d(d(z0))), D(d(z0)), D(z0))
C(z0) → c3(G(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
D(d(d(z0))) → c5(G(c(z0)), C(z0))
F(z0) → c6(G(z0))
G(z0) → c7(D(a(b(z0))), B(z0))
G(g(z0)) → c8(B(c(z0)), C(z0))

K tuples:none
Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

B, C, D, F, G

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7, c8

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

C(c(z0)) → c2(D(d(d(z0))), D(d(z0)), D(z0))
D(d(d(z0))) → c5(G(c(z0)), C(z0))
G(g(z0)) → c8(B(c(z0)), C(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
G(z0) → c7(D(a(b(z0))), B(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))

S tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(z0) → c3(G(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
F(z0) → c6(G(z0))
G(z0) → c7(D(a(b(z0))), B(z0))

K tuples:none
Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

C, F, G, B, D

Compound Symbols:

c3, c6, c7, c1, c4

(5) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

S tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(z0) → c3(G(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
F(z0) → c6(G(z0))
G(z0) → c7(B(z0))

K tuples:none
Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

C, F, B, D, G

Compound Symbols:

c3, c6, c1, c4, c7

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

D(d(z0)) → c4(C(f(z0)), F(z0))

We considered the (Usable) Rules:

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

And the Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(B(x₁)) = [2]x₁   
POL(C(x₁)) = [2]x₁   
POL(D(x₁)) = [4] + [4]x₁   
POL(F(x₁)) = [4]x₁   
POL(G(x₁)) = [2]x₁   
POL(a(x₁)) = 0   
POL(b(x₁)) = [4] + [5]x₁   
POL(c(x₁)) = [3] + [5]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(d(x₁)) = [2] + [2]x₁   
POL(f(x₁)) = 0   
POL(g(x₁)) = [3] + [3]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

S tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
G(z0) → c7(B(z0))

K tuples:

D(d(z0)) → c4(C(f(z0)), F(z0))

Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

C, F, B, D, G

Compound Symbols:

c3, c6, c1, c4, c7

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(z0) → c6(G(z0))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

S tuples:

B(b(z0)) → c1(C(d(z0)), D(z0))
C(z0) → c3(G(z0))
G(z0) → c7(B(z0))

K tuples:

D(d(z0)) → c4(C(f(z0)), F(z0))
F(z0) → c6(G(z0))

Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

C, F, B, D, G

Compound Symbols:

c3, c6, c1, c4, c7

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

B(b(z0)) → c1(C(d(z0)), D(z0))

We considered the (Usable) Rules:

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

And the Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(B(x₁)) = [2]x₁   
POL(C(x₁)) = [4]x₁   
POL(D(x₁)) = [2]x₁   
POL(F(x₁)) = [4]x₁   
POL(G(x₁)) = [4]x₁   
POL(a(x₁)) = 0   
POL(b(x₁)) = [4] + [5]x₁   
POL(c(x₁)) = [2]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(d(x₁)) = [2]x₁   
POL(f(x₁)) = 0   
POL(g(x₁)) = [2]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

C(z0) → c3(G(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))
D(d(z0)) → c4(C(f(z0)), F(z0))
G(z0) → c7(B(z0))

S tuples:

C(z0) → c3(G(z0))
G(z0) → c7(B(z0))

K tuples:

D(d(z0)) → c4(C(f(z0)), F(z0))
F(z0) → c6(G(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))

Defined Rule Symbols:

b, c, d, f, g

Defined Pair Symbols:

C, F, B, D, G

Compound Symbols:

c3, c6, c1, c4, c7

(13) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

C(z0) → c3(G(z0))
G(z0) → c7(B(z0))
G(z0) → c7(B(z0))
B(b(z0)) → c1(C(d(z0)), D(z0))

Now S is empty

(14) BOUNDS(O(1), O(1))