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:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → if(z0, c, n__f(true))
f(z0) → n__f(z0)
if(true, z0, z1) → z0
if(false, z0, z1) → activate(z1)
activate(n__f(z0)) → f(z0)
activate(z0) → z0

Tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
F(z0) → c2
IF(true, z0, z1) → c3
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))
ACTIVATE(z0) → c6

S tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
F(z0) → c2
IF(true, z0, z1) → c3
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))
ACTIVATE(z0) → c6

K tuples:none
Defined Rule Symbols:

f, if, activate

Defined Pair Symbols:

F, IF, ACTIVATE

Compound Symbols:

c1, c2, c3, c4, c5, c6

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

Removed 3 trailing nodes:

F(z0) → c2
ACTIVATE(z0) → c6
IF(true, z0, z1) → c3

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → if(z0, c, n__f(true))
f(z0) → n__f(z0)
if(true, z0, z1) → z0
if(false, z0, z1) → activate(z1)
activate(n__f(z0)) → f(z0)
activate(z0) → z0

Tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

S tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

K tuples:none
Defined Rule Symbols:

f, if, activate

Defined Pair Symbols:

F, IF, ACTIVATE

Compound Symbols:

c1, c4, c5

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0) → if(z0, c, n__f(true))
f(z0) → n__f(z0)
if(true, z0, z1) → z0
if(false, z0, z1) → activate(z1)
activate(n__f(z0)) → f(z0)
activate(z0) → z0

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

S tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F, IF, ACTIVATE

Compound Symbols:

c1, c4, c5

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

IF(false, z0, z1) → c4(ACTIVATE(z1))

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

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

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

POL(ACTIVATE(x₁)) = [2]x₁   
POL(F(x₁)) = [2]x₁   
POL(IF(x₁, x₂, x₃)) = [2]x₁ + [4]x₃   
POL(c) = [2]   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(false) = [3]   
POL(n__f(x₁)) = x₁   
POL(true) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))
ACTIVATE(n__f(z0)) → c5(F(z0))

S tuples:

F(z0) → c1(IF(z0, c, n__f(true)))
ACTIVATE(n__f(z0)) → c5(F(z0))

K tuples:

IF(false, z0, z1) → c4(ACTIVATE(z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F, IF, ACTIVATE

Compound Symbols:

c1, c4, c5

(9) CdtKnowledgeProof (EQUIVALENT transformation)

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

ACTIVATE(n__f(z0)) → c5(F(z0))
F(z0) → c1(IF(z0, c, n__f(true)))
IF(false, z0, z1) → c4(ACTIVATE(z1))

Now S is empty

(10) BOUNDS(O(1), O(1))