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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(a(z0))) → c1(B(z0))
B(b(b(z0))) → c2(C(a(c(z0))), C(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))

S tuples:

B(c(a(z0))) → c1(B(z0))
B(b(b(z0))) → c2(C(a(c(z0))), C(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))

K tuples:none
Defined Rule Symbols:

b, c, d

Defined Pair Symbols:

B, C, D

Compound Symbols:

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

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

S tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

K tuples:none
Defined Rule Symbols:

b, c, d

Defined Pair Symbols:

B, C, D

Compound Symbols:

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

(5) 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(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
B(b(b(z0))) → c2(C(z0))

We considered the (Usable) Rules:

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

And the Tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

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

POL(B(x₁)) = x₁   
POL(C(x₁)) = [4]x₁   
POL(D(x₁)) = [4]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [1] + [2]x₁   
POL(c(x₁)) = [1] + [2]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(d(x₁)) = [3] + [4]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

S tuples:

C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))

K tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
B(b(b(z0))) → c2(C(z0))

Defined Rule Symbols:

b, c, d

Defined Pair Symbols:

B, C, D

Compound Symbols:

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

(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.

C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))

We considered the (Usable) Rules:

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

And the Tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

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

POL(B(x₁)) = x₁   
POL(C(x₁)) = [2] + [4]x₁   
POL(D(x₁)) = [3] + [4]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [1] + [2]x₁   
POL(c(x₁)) = [1] + [2]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(d(x₁)) = [3] + [4]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(b(b(z0))) → c2(C(z0))

S tuples:

D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))

K tuples:

B(c(a(z0))) → c1(B(z0))
C(d(z0)) → c3(D(c(z0)), C(z0))
C(b(z0)) → c5(D(a(z0)))
B(b(b(z0))) → c2(C(z0))
C(d(b(z0))) → c4(D(c(c(z0))), C(c(z0)), C(z0))

Defined Rule Symbols:

b, c, d

Defined Pair Symbols:

B, C, D

Compound Symbols:

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

(9) CdtKnowledgeProof (EQUIVALENT transformation)

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

D(c(z0)) → c6(B(b(b(z0))), B(b(z0)), B(z0))
D(a(z0)) → c8(B(z0))
B(c(a(z0))) → c1(B(z0))
B(b(b(z0))) → c2(C(z0))
B(c(a(z0))) → c1(B(z0))
B(b(b(z0))) → c2(C(z0))

Now S is empty

(10) BOUNDS(O(1), O(1))