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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
↳2 CdtProblem
↳3 CdtUnreachableProof (⇔)
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
↳6 CdtProblem
↳7 CdtKnowledgeProof (⇔)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
↳10 CdtProblem
↳11 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
↳12 CdtProblem
↳13 CdtUnreachableProof (⇔)
↳14 CdtProblem
↳15 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
↳16 CdtProblem
↳17 CdtUnreachableProof (⇔)
↳18 CdtProblem
↳19 CdtGraphRemoveDanglingProof (BOTH BOUNDS(ID, ID))
↳20 CdtProblem
↳21 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
↳22 CdtProblem
↳23 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
↳24 CdtProblem
↳25 SIsEmptyProof (⇔)
↳26 BOUNDS(O(1), O(1))

(0) Obligation:

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

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → c(x1)
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
C(f(z0)) → c2(F(n(a(c(z0)))), N(a(c(z0))), C(z0))
C(c(z0)) → c3(C(z0))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
N(f(z0)) → c6(F(n(z0)), N(z0))
S tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
C(f(z0)) → c2(F(n(a(c(z0)))), N(a(c(z0))), C(z0))
C(c(z0)) → c3(C(z0))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
N(f(z0)) → c6(F(n(z0)), N(z0))
K tuples:none
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

F, C, N

Compound Symbols:

c1, c2, c3, c4, c5, c6

(3) CdtUnreachableProof (EQUIVALENT transformation)

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

C(f(z0)) → c2(F(n(a(c(z0)))), N(a(c(z0))), C(z0))
N(f(z0)) → c6(F(n(z0)), N(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
S tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
C(c(z0)) → c3(C(z0))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
K tuples:none
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

F, N, C

Compound Symbols:

c1, c4, c5, c3

(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.

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
We considered the (Usable) Rules:

n(a(z0)) → c(z0)
c(c(z0)) → c(z0)
And the Tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(C(x1)) = 0   
POL(F(x1)) = [1]   
POL(N(x1)) = [2]x1   
POL(a(x1)) = 0   
POL(c(x1)) = 0   
POL(c1(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(n(x1)) = [4]   
POL(s(x1)) = [3]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
S tuples:

C(c(z0)) → c3(C(z0))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
K tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

F, N, C

Compound Symbols:

c1, c4, c5, c3

(7) CdtKnowledgeProof (EQUIVALENT transformation)

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

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

F, N, C

Compound Symbols:

c1, c4, c5, c3

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

Use narrowing to replace F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0))) by

F(z0) → c1(N(c(c(z0))), C(n(a(z0))), N(a(z0)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
F(z0) → c1(N(c(c(z0))), C(n(a(z0))), N(a(z0)))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, C, F

Compound Symbols:

c4, c5, c3, c1

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

Use narrowing to replace F(z0) → c1(N(c(c(z0))), C(n(a(z0))), N(a(z0))) by

F(z0) → c1(N(c(z0)), C(n(a(z0))), N(a(z0)))
F(c(z0)) → c1(N(c(c(z0))), C(n(a(c(z0)))), N(a(c(z0))))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
F(z0) → c1(N(c(z0)), C(n(a(z0))), N(a(z0)))
F(c(z0)) → c1(N(c(c(z0))), C(n(a(c(z0)))), N(a(c(z0))))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

F(z0) → c1(N(c(n(a(z0)))), C(n(a(z0))), N(a(z0)))
N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, C, F

Compound Symbols:

c4, c5, c3, c1

(13) CdtUnreachableProof (EQUIVALENT transformation)

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

F(c(z0)) → c1(N(c(c(z0))), C(n(a(c(z0)))), N(a(c(z0))))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
F(z0) → c1(N(c(z0)), C(n(a(z0))), N(a(z0)))
C(c(z0)) → c3(C(z0))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, F, C

Compound Symbols:

c4, c5, c1, c3

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

Use narrowing to replace F(z0) → c1(N(c(z0)), C(n(a(z0))), N(a(z0))) by

F(c(z0)) → c1(N(c(z0)), C(n(a(c(z0)))), N(a(c(z0))))
F(x0) → c1(C(n(a(x0))), N(a(x0)))

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
C(c(z0)) → c3(C(z0))
F(c(z0)) → c1(N(c(z0)), C(n(a(c(z0)))), N(a(c(z0))))
F(x0) → c1(C(n(a(x0))), N(a(x0)))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, C, F

Compound Symbols:

c4, c5, c3, c1, c1

(17) CdtUnreachableProof (EQUIVALENT transformation)

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

F(c(z0)) → c1(N(c(z0)), C(n(a(c(z0)))), N(a(c(z0))))

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
F(x0) → c1(C(n(a(x0))), N(a(x0)))
C(c(z0)) → c3(C(z0))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

N(a(z0)) → c4(C(z0))
N(s(z0)) → c5(F(s(s(z0))))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, F, C

Compound Symbols:

c4, c5, c1, c3

(19) CdtGraphRemoveDanglingProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 of 4 dangling nodes:

N(s(z0)) → c5(F(s(s(z0))))

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
F(x0) → c1(C(n(a(x0))), N(a(x0)))
C(c(z0)) → c3(C(z0))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

N(a(z0)) → c4(C(z0))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, F, C

Compound Symbols:

c4, c1, c3

(21) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
C(c(z0)) → c3(C(z0))
F(x0) → c2(C(n(a(x0))))
F(x0) → c2(N(a(x0)))
S tuples:

C(c(z0)) → c3(C(z0))
K tuples:

N(a(z0)) → c4(C(z0))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, C, F

Compound Symbols:

c4, c3, c2

(23) 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(c(z0)) → c3(C(z0))
We considered the (Usable) Rules:

n(a(z0)) → c(z0)
c(c(z0)) → c(z0)
And the Tuples:

N(a(z0)) → c4(C(z0))
C(c(z0)) → c3(C(z0))
F(x0) → c2(C(n(a(x0))))
F(x0) → c2(N(a(x0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(C(x1)) = [2]x1   
POL(F(x1)) = [2] + [3]x1 + x12   
POL(N(x1)) = [3]x1 + x12   
POL(a(x1)) = x1   
POL(c(x1)) = [1] + x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(n(x1)) = [1] + x1   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → n(c(n(a(z0))))
c(f(z0)) → f(n(a(c(z0))))
c(c(z0)) → c(z0)
n(a(z0)) → c(z0)
n(s(z0)) → f(s(s(z0)))
n(f(z0)) → f(n(z0))
Tuples:

N(a(z0)) → c4(C(z0))
C(c(z0)) → c3(C(z0))
F(x0) → c2(C(n(a(x0))))
F(x0) → c2(N(a(x0)))
S tuples:none
K tuples:

N(a(z0)) → c4(C(z0))
C(c(z0)) → c3(C(z0))
Defined Rule Symbols:

f, c, n

Defined Pair Symbols:

N, C, F

Compound Symbols:

c4, c3, c2

(25) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(26) BOUNDS(O(1), O(1))