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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(a(b(z0))) → c2(A(d(b(z0))), D(b(z0)), B(z0))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
D(d(z0)) → c5(A(d(b(z0))), D(b(z0)), B(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(a(b(z0))) → c2(A(d(b(z0))), D(b(z0)), B(z0))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
D(d(z0)) → c5(A(d(b(z0))), D(b(z0)), B(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

K tuples:none
Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

A, D, B

Compound Symbols:

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

(3) CdtUnreachableProof (EQUIVALENT transformation)

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

A(a(b(z0))) → c2(A(d(b(z0))), D(b(z0)), B(z0))
D(d(z0)) → c5(A(d(b(z0))), D(b(z0)), B(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

K tuples:none
Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

B, A

Compound Symbols:

c7, c1, c3, c4, c6

(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.

A(d(c(z0))) → c4(A(z0))

We considered the (Usable) Rules:

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

And the Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

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

POL(A(x₁)) = x₁   
POL(B(x₁)) = [3]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [5]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(d(x₁)) = [2] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

K tuples:

A(d(c(z0))) → c4(A(z0))

Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

B, A

Compound Symbols:

c7, c1, c3, c4, c6

(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.

B(b(z0)) → c6(B(c(z0)))

We considered the (Usable) Rules:

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

And the Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

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

POL(A(x₁)) = x₁   
POL(B(x₁)) = [3]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [4] + [5]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(d(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

K tuples:

A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

B, A

Compound Symbols:

c7, c1, c3, c4, c6

(9) 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(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

We considered the (Usable) Rules:

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

And the Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

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

POL(A(x₁)) = x₁   
POL(B(x₁)) = [4] + [3]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [4] + [5]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(d(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

S tuples:

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

K tuples:

A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))

Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

B, A

Compound Symbols:

c7, c1, c3, c4, c6

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

We considered the (Usable) Rules:

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

And the Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

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

POL(A(x₁)) = x₁   
POL(B(x₁)) = [3]x₁   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [1] + [5]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(d(x₁)) = [4]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))

K tuples:

A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))

Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

B, A

Compound Symbols:

c7, c1, c3, c4, c6

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

Use narrowing to replace B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0)) by

B(c(c(a(z0)))) → c7(A(a(c(a(c(z0))))), A(a(c(a(z0)))), A(c(a(z0))))
B(c(b(z0))) → c7(A(a(b(a(a(z0))))), A(a(b(z0))), A(b(z0)))
B(c(d(c(z0)))) → c7(A(a(c(a(z0)))), A(a(d(c(z0)))), A(d(c(z0))))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(c(a(z0)))) → c7(A(a(c(a(c(z0))))), A(a(c(a(z0)))), A(c(a(z0))))
B(c(b(z0))) → c7(A(a(b(a(a(z0))))), A(a(b(z0))), A(b(z0)))
B(c(d(c(z0)))) → c7(A(a(c(a(z0)))), A(a(d(c(z0)))), A(d(c(z0))))

S tuples:

A(c(a(z0))) → c1(A(c(z0)))

K tuples:

A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(z0)) → c7(A(a(a(z0))), A(a(z0)), A(z0))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))

Defined Rule Symbols:

a, d, b

Defined Pair Symbols:

A, B

Compound Symbols:

c1, c3, c4, c6, c7

(15) CdtUnreachableProof (EQUIVALENT transformation)

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

A(c(a(z0))) → c1(A(c(z0)))
A(b(z0)) → c3(B(a(a(z0))), A(a(z0)), A(z0))
A(d(c(z0))) → c4(A(z0))
B(b(z0)) → c6(B(c(z0)))
B(c(c(a(z0)))) → c7(A(a(c(a(c(z0))))), A(a(c(a(z0)))), A(c(a(z0))))
B(c(b(z0))) → c7(A(a(b(a(a(z0))))), A(a(b(z0))), A(b(z0)))
B(c(d(c(z0)))) → c7(A(a(c(a(z0)))), A(a(d(c(z0)))), A(d(c(z0))))

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

a, d, b

Defined Pair Symbols:none

Compound Symbols:none

(17) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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