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:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(z0)) → c1(B(a(z0)), A(z0))
B(c(z0)) → c3(A(b(z0)), B(z0))

S tuples:

A(d(z0)) → c1(B(a(z0)), A(z0))
B(c(z0)) → c3(A(b(z0)), B(z0))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

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

Use narrowing to replace A(d(z0)) → c1(B(a(z0)), A(z0)) by

A(d(d(z0))) → c1(B(d(c(b(a(z0))))), A(d(z0)))
A(d(c(z0))) → c1(B(z0), A(c(z0)))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c3(A(b(z0)), B(z0))
A(d(d(z0))) → c1(B(d(c(b(a(z0))))), A(d(z0)))
A(d(c(z0))) → c1(B(z0), A(c(z0)))

S tuples:

B(c(z0)) → c3(A(b(z0)), B(z0))
A(d(d(z0))) → c1(B(d(c(b(a(z0))))), A(d(z0)))
A(d(c(z0))) → c1(B(z0), A(c(z0)))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

B, A

Compound Symbols:

c3, c1

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

Removed 2 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c3(A(b(z0)), B(z0))
A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))

S tuples:

B(c(z0)) → c3(A(b(z0)), B(z0))
A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

B, A

Compound Symbols:

c3, c1

(7) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace B(c(z0)) → c3(A(b(z0)), B(z0)) by

B(c(c(z0))) → c3(A(c(d(a(b(z0))))), B(c(z0)))
B(c(d(z0))) → c3(A(z0), B(d(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(A(c(d(a(b(z0))))), B(c(z0)))
B(c(d(z0))) → c3(A(z0), B(d(z0)))

S tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(A(c(d(a(b(z0))))), B(c(z0)))
B(c(d(z0))) → c3(A(z0), B(d(z0)))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

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

Removed 2 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

S tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

(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.

A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))

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

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

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

POL(A(x₁)) = [2] + [4]x₁   
POL(B(x₁)) = [4]x₁   
POL(c(x₁)) = [4] + x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(d(x₁)) = [2] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

S tuples:

A(d(d(z0))) → c1(A(d(z0)))
B(c(d(z0))) → c3(A(z0))

K tuples:

A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))

Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

(13) CdtKnowledgeProof (EQUIVALENT transformation)

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

B(c(d(z0))) → c3(A(z0))
A(d(c(z0))) → c1(B(z0))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

S tuples:

A(d(d(z0))) → c1(A(d(z0)))

K tuples:

A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

(15) 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.

A(d(d(z0))) → c1(A(d(z0)))

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

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

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

POL(A(x₁)) = [1] + [4]x₁   
POL(B(x₁)) = [4] + [4]x₁   
POL(c(x₁)) = [4] + x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(d(x₁)) = [4] + x₁   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(d(d(z0))) → c1(A(d(z0)))
A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))

S tuples:none
K tuples:

A(d(c(z0))) → c1(B(z0))
B(c(c(z0))) → c3(B(c(z0)))
B(c(d(z0))) → c3(A(z0))
A(d(d(z0))) → c1(A(d(z0)))

Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3

(17) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(18) BOUNDS(O(1), O(1))