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:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(a) → c3(ACTIVE(a))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(a) → c3(ACTIVE(a))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:none
Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

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

Removed 1 trailing nodes:

MARK(a) → c3(ACTIVE(a))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:none
Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(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.

C(active(z0)) → c9(C(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(C(x₁)) = x₁   
POL(F(x₁)) = 0   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = x₁   
POL(a) = 0   
POL(active(x₁)) = [1] + [2]x₁   
POL(c(x₁)) = [4]x₁   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [2]x₁   
POL(g(x₁)) = x₁   
POL(mark(x₁)) = [4]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(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.

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(C(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = [2]x₁   
POL(a) = [2]   
POL(active(x₁)) = x₁   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [2]x₁   
POL(g(x₁)) = [4]x₁   
POL(mark(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(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.

MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(C(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = x₁   
POL(MARK(x₁)) = x₁   
POL(a) = 0   
POL(active(x₁)) = x₁   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [4]x₁   
POL(g(x₁)) = [1] + [4]x₁   
POL(mark(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(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.

MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(C(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = x₁   
POL(a) = 0   
POL(active(x₁)) = [1]   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [4] + [2]x₁   
POL(g(x₁)) = [4]x₁   
POL(mark(x₁)) = [1] + [4]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(13) CdtKnowledgeProof (EQUIVALENT transformation)

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

MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(active(z0)) → c9(C(z0))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(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.

C(mark(z0)) → c8(C(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(C(x₁)) = x₁   
POL(F(x₁)) = 0   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = x₁   
POL(a) = 0   
POL(active(x₁)) = [2]x₁   
POL(c(x₁)) = [4]x₁   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [2]x₁   
POL(g(x₁)) = [2]x₁   
POL(mark(x₁)) = [1] + [4]x₁   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(mark(z0)) → c8(C(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(17) 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(mark(z0)) → c6(F(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(C(x₁)) = 0   
POL(F(x₁)) = [1] + [2]x₁   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = [4]x₁   
POL(a) = 0   
POL(active(x₁)) = x₁   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [4] + [4]x₁   
POL(g(x₁)) = [1] + [2]x₁   
POL(mark(x₁)) = [1] + [2]x₁   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

F(active(z0)) → c7(F(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(mark(z0)) → c8(C(z0))
F(mark(z0)) → c6(F(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(19) 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.

G(mark(z0)) → c10(G(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(C(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = [2]x₁   
POL(MARK(x₁)) = [4]x₁   
POL(a) = 0   
POL(active(x₁)) = x₁   
POL(c(x₁)) = 0   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [1] + [2]x₁   
POL(g(x₁)) = [4] + [4]x₁   
POL(mark(x₁)) = [1] + [2]x₁   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

F(active(z0)) → c7(F(z0))
G(active(z0)) → c11(G(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(mark(z0)) → c8(C(z0))
F(mark(z0)) → c6(F(z0))
G(mark(z0)) → c10(G(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(21) 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.

G(active(z0)) → c11(G(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(C(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = x₁   
POL(MARK(x₁)) = [4]x₁   
POL(a) = [1]   
POL(active(x₁)) = [1] + x₁   
POL(c(x₁)) = [1]   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [1] + [2]x₁   
POL(g(x₁)) = [1] + [5]x₁   
POL(mark(x₁)) = [2]x₁   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:

F(active(z0)) → c7(F(z0))

K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(mark(z0)) → c8(C(z0))
F(mark(z0)) → c6(F(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(23) 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(active(z0)) → c7(F(z0))

We considered the (Usable) Rules:

mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
g(active(z0)) → g(z0)
g(mark(z0)) → g(z0)
active(f(f(a))) → mark(c(f(g(f(a)))))
f(active(z0)) → f(z0)
f(mark(z0)) → f(z0)

And the Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(C(x₁)) = [4]   
POL(F(x₁)) = x₁   
POL(G(x₁)) = [2]   
POL(MARK(x₁)) = [5] + [4]x₁   
POL(a) = [4]   
POL(active(x₁)) = [1] + x₁   
POL(c(x₁)) = [4]   
POL(c1(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(f(x₁)) = [1] + [4]x₁   
POL(g(x₁)) = [1] + [2]x₁   
POL(mark(x₁)) = [2]x₁   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(z0)) → active(f(mark(z0)))
mark(a) → active(a)
mark(c(z0)) → active(c(z0))
mark(g(z0)) → active(g(mark(z0)))
f(mark(z0)) → f(z0)
f(active(z0)) → f(z0)
c(mark(z0)) → c(z0)
c(active(z0)) → c(z0)
g(mark(z0)) → g(z0)
g(active(z0)) → g(z0)

Tuples:

ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
F(mark(z0)) → c6(F(z0))
F(active(z0)) → c7(F(z0))
C(mark(z0)) → c8(C(z0))
C(active(z0)) → c9(C(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))

S tuples:none
K tuples:

C(active(z0)) → c9(C(z0))
ACTIVE(f(f(a))) → c1(MARK(c(f(g(f(a))))), C(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(g(z0)) → c5(ACTIVE(g(mark(z0))), G(mark(z0)), MARK(z0))
MARK(f(z0)) → c2(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(c(z0)) → c4(ACTIVE(c(z0)), C(z0))
C(mark(z0)) → c8(C(z0))
F(mark(z0)) → c6(F(z0))
G(mark(z0)) → c10(G(z0))
G(active(z0)) → c11(G(z0))
F(active(z0)) → c7(F(z0))

Defined Rule Symbols:

active, mark, f, c, g

Defined Pair Symbols:

ACTIVE, MARK, F, C, G

Compound Symbols:

c1, c2, c4, c5, c6, c7, c8, c9, c10, c11

(25) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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