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

(0) Obligation:

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))

Tuples:

F(0) → c1
F(1) → c2
F(s(z0)) → c3(F(z0))
IF(true, s(z0), s(z1)) → c4
IF(false, s(z0), s(z1)) → c5
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))

S tuples:

F(0) → c1
F(1) → c2
F(s(z0)) → c3(F(z0))
IF(true, s(z0), s(z1)) → c4
IF(false, s(z0), s(z1)) → c5
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))

K tuples:none
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, IF, G

Compound Symbols:

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

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

Removed 4 trailing nodes:

F(1) → c2
IF(false, s(z0), s(z1)) → c5
F(0) → c1
IF(true, s(z0), s(z1)) → c4

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))

Tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))

S tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))

K tuples:none
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))

Tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

S tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

K tuples:none
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

S tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

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

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

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

POL(F(x₁)) = [4]   
POL(G(x₁, x₂)) = [4]x₁ + [2]x₂   
POL(c(x₁)) = [4] + x₁   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

S tuples:

F(s(z0)) → c3(F(z0))

K tuples:

G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

F(s(z0)) → c3(F(z0))

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

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

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

POL(F(x₁)) = x₁   
POL(G(x₁, x₂)) = x₂² + x₁·x₂   
POL(c(x₁)) = [1] + x₁   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))

S tuples:none
K tuples:

G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
F(s(z0)) → c3(F(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

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

The set S is empty

(14) BOUNDS(1, 1)