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, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
++(x, nil) → x
++(x, g(y, z)) → g(++(x, y), z)
null(nil) → true
null(g(x, y)) → false
mem(nil, y) → false
mem(g(x, y), z) → or(=(y, z), mem(x, z))
mem(x, max(x)) → not(null(x))
max(g(g(nil, x), y)) → max'(x, y)
max(g(g(g(x, y), z), u)) → max'(max(g(g(x, y), z)), u)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
++(z0, nil) → z0
++(z0, g(z1, z2)) → g(++(z0, z1), z2)
null(nil) → true
null(g(z0, z1)) → false
mem(nil, z0) → false
mem(g(z0, z1), z2) → or(=(z1, z2), mem(z0, z2))
mem(z0, max(z0)) → not(null(z0))
max(g(g(nil, z0), z1)) → max'(z0, z1)
max(g(g(g(z0, z1), z2), u)) → max'(max(g(g(z0, z1), z2)), u)

Tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MEM(z0, max(z0)) → c8(NULL(z0))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

S tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MEM(z0, max(z0)) → c8(NULL(z0))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

K tuples:none
Defined Rule Symbols:

f, ++, null, mem, max

Defined Pair Symbols:

F, ++', MEM, MAX

Compound Symbols:

c1, c3, c7, c8, c10

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

Removed 1 trailing nodes:

MEM(z0, max(z0)) → c8(NULL(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
++(z0, nil) → z0
++(z0, g(z1, z2)) → g(++(z0, z1), z2)
null(nil) → true
null(g(z0, z1)) → false
mem(nil, z0) → false
mem(g(z0, z1), z2) → or(=(z1, z2), mem(z0, z2))
mem(z0, max(z0)) → not(null(z0))
max(g(g(nil, z0), z1)) → max'(z0, z1)
max(g(g(g(z0, z1), z2), u)) → max'(max(g(g(z0, z1), z2)), u)

Tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

S tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

K tuples:none
Defined Rule Symbols:

f, ++, null, mem, max

Defined Pair Symbols:

F, ++', MEM, MAX

Compound Symbols:

c1, c3, c7, c10

(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, g(z1, z2)) → c1(F(z0, z1))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

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

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

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

POL(++'(x₁, x₂)) = 0   
POL(F(x₁, x₂)) = x₂   
POL(MAX(x₁)) = x₁   
POL(MEM(x₁, x₂)) = 0   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(g(x₁, x₂)) = [1] + x₁   
POL(u) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
++(z0, nil) → z0
++(z0, g(z1, z2)) → g(++(z0, z1), z2)
null(nil) → true
null(g(z0, z1)) → false
mem(nil, z0) → false
mem(g(z0, z1), z2) → or(=(z1, z2), mem(z0, z2))
mem(z0, max(z0)) → not(null(z0))
max(g(g(nil, z0), z1)) → max'(z0, z1)
max(g(g(g(z0, z1), z2), u)) → max'(max(g(g(z0, z1), z2)), u)

Tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

S tuples:

++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))

K tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

Defined Rule Symbols:

f, ++, null, mem, max

Defined Pair Symbols:

F, ++', MEM, MAX

Compound Symbols:

c1, c3, c7, c10

(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.

++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))

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

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

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

POL(++'(x₁, x₂)) = [5]x₂   
POL(F(x₁, x₂)) = 0   
POL(MAX(x₁)) = 0   
POL(MEM(x₁, x₂)) = [4]x₁   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(g(x₁, x₂)) = [4] + x₁   
POL(u) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
++(z0, nil) → z0
++(z0, g(z1, z2)) → g(++(z0, z1), z2)
null(nil) → true
null(g(z0, z1)) → false
mem(nil, z0) → false
mem(g(z0, z1), z2) → or(=(z1, z2), mem(z0, z2))
mem(z0, max(z0)) → not(null(z0))
max(g(g(nil, z0), z1)) → max'(z0, z1)
max(g(g(g(z0, z1), z2), u)) → max'(max(g(g(z0, z1), z2)), u)

Tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))

S tuples:none
K tuples:

F(z0, g(z1, z2)) → c1(F(z0, z1))
MAX(g(g(g(z0, z1), z2), u)) → c10(MAX(g(g(z0, z1), z2)))
++'(z0, g(z1, z2)) → c3(++'(z0, z1))
MEM(g(z0, z1), z2) → c7(MEM(z0, z2))

Defined Rule Symbols:

f, ++, null, mem, max

Defined Pair Symbols:

F, ++', MEM, MAX

Compound Symbols:

c1, c3, c7, c10

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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