The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(a(z0)) → c1(B(c(c(c(z0)))), C(c(c(z0))), C(c(z0)), C(z0))
A(z0) → c2(C(d(z0)))
B(b(z0)) → c4(C(c(c(z0))), C(c(z0)), C(z0))
C(d(d(z0))) → c6(A(z0))

S tuples:

A(a(z0)) → c1(B(c(c(c(z0)))), C(c(c(z0))), C(c(z0)), C(z0))
A(z0) → c2(C(d(z0)))
B(b(z0)) → c4(C(c(c(z0))), C(c(z0)), C(z0))
C(d(d(z0))) → c6(A(z0))

K tuples:none
Defined Rule Symbols:

a, b, c

Defined Pair Symbols:

A, B, C

Compound Symbols:

c1, c2, c4, c6

(3) CdtUnreachableProof (EQUIVALENT transformation)

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

A(a(z0)) → c1(B(c(c(c(z0)))), C(c(c(z0))), C(c(z0)), C(z0))
B(b(z0)) → c4(C(c(c(z0))), C(c(z0)), C(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(z0) → c2(C(d(z0)))
C(d(d(z0))) → c6(A(z0))

S tuples:

A(z0) → c2(C(d(z0)))
C(d(d(z0))) → c6(A(z0))

K tuples:none
Defined Rule Symbols:

a, b, c

Defined Pair Symbols:

A, C

Compound Symbols:

c2, c6

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

C(d(d(z0))) → c6(A(z0))

We considered the (Usable) Rules:none
And the Tuples:

A(z0) → c2(C(d(z0)))
C(d(d(z0))) → c6(A(z0))

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

POL(A(x₁)) = [2] + x₁   
POL(C(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(d(x₁)) = [2] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(z0) → c2(C(d(z0)))
C(d(d(z0))) → c6(A(z0))

S tuples:

A(z0) → c2(C(d(z0)))

K tuples:

C(d(d(z0))) → c6(A(z0))

Defined Rule Symbols:

a, b, c

Defined Pair Symbols:

A, C

Compound Symbols:

c2, c6

(7) CdtKnowledgeProof (EQUIVALENT transformation)

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

A(z0) → c2(C(d(z0)))
C(d(d(z0))) → c6(A(z0))

Now S is empty

(8) BOUNDS(O(1), O(1))