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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

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

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

Removed 1 trailing nodes:

MARK(a) → c2(ACTIVE(a))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

K tuples:none
Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(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.

G(mark(z0)) → c6(G(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = [2]x₁   
POL(MARK(x₁)) = [2]x₁   
POL(a) = 0   
POL(active(x₁)) = x₁   
POL(c(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(f(x₁)) = x₁   
POL(g(x₁)) = x₁   
POL(mark(x₁)) = [4] + [2]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(active(z0)) → c7(G(z0))

K tuples:

G(mark(z0)) → c6(G(z0))

Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(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.

G(active(z0)) → c7(G(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

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

POL(ACTIVE(x₁)) = 0   
POL(F(x₁)) = 0   
POL(G(x₁)) = x₁   
POL(MARK(x₁)) = [2]x₁   
POL(a) = 0   
POL(active(x₁)) = [1] + x₁   
POL(c(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(f(x₁)) = x₁   
POL(g(x₁)) = x₁   
POL(mark(x₁)) = [2]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))

K tuples:

G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(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.

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

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

POL(ACTIVE(x₁)) = [2]x₁   
POL(F(x₁)) = [1] + [4]x₁   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = [1] + [4]x₁   
POL(a) = 0   
POL(active(x₁)) = x₁   
POL(c(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(f(x₁)) = [2] + [4]x₁   
POL(g(x₁)) = 0   
POL(mark(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:

F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))

K tuples:

G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))
ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))

Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(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.

F(active(z0)) → c5(F(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

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

POL(ACTIVE(x₁)) = [2] + x₁   
POL(F(x₁)) = x₁   
POL(G(x₁)) = 0   
POL(MARK(x₁)) = [4]x₁   
POL(a) = [1]   
POL(active(x₁)) = [1] + x₁   
POL(c(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(f(x₁)) = [1] + [4]x₁   
POL(g(x₁)) = [1]   
POL(mark(x₁)) = [2]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:

F(mark(z0)) → c4(F(z0))

K tuples:

G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))
ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(active(z0)) → c5(F(z0))

Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(13) 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)) → c4(F(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

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

POL(ACTIVE(x₁)) = x₁   
POL(F(x₁)) = [2] + [2]x₁   
POL(G(x₁)) = [4]   
POL(MARK(x₁)) = [4] + [4]x₁   
POL(a) = [1]   
POL(active(x₁)) = x₁   
POL(c(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(f(x₁)) = [3] + [4]x₁   
POL(g(x₁)) = 0   
POL(mark(x₁)) = [1] + [2]x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(mark(z0)) → c4(F(z0))
F(active(z0)) → c5(F(z0))
G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))

S tuples:none
K tuples:

G(mark(z0)) → c6(G(z0))
G(active(z0)) → c7(G(z0))
ACTIVE(f(f(a))) → c(MARK(f(g(f(a)))), F(g(f(a))), G(f(a)), F(a))
MARK(f(z0)) → c1(ACTIVE(f(mark(z0))), F(mark(z0)), MARK(z0))
MARK(g(z0)) → c3(ACTIVE(g(z0)), G(z0))
F(active(z0)) → c5(F(z0))
F(mark(z0)) → c4(F(z0))

Defined Rule Symbols:

active, mark, f, g

Defined Pair Symbols:

ACTIVE, MARK, F, G

Compound Symbols:

c, c1, c3, c4, c5, c6, c7

(15) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(16) BOUNDS(O(1), O(1))