Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
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)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, g(y, z)) → F(x, y)
++¹(x, g(y, z)) → ++¹(x, y)
MEM(g(x, y), z) → MEM(x, z)
MEM(x, max(x)) → NULL(x)
MAX(g(g(g(x, y), z), u)) → MAX(g(g(x, y), z))

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MAX(g(g(g(x, y), z), u)) → MAX(g(g(x, y), z))

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MAX(g(g(g(x, y), z), u)) → MAX(g(g(x, y), z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAX(x1)  =  x1
g(x1, x2)  =  g(x1, x2)
u  =  u
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null
true  =  true
false  =  false
mem(x1, x2)  =  mem
or(x1, x2)  =  or
=(x1, x2)  =  =(x1, x2)
max(x1)  =  max
not(x1)  =  not(x1)
max'(x1, x2)  =  max'

Lexicographic path order with status [LPO].
Precedence:

u > g₂ > max'
f₂ > g₂ > max'
nil > g₂ > max'
nil > false > max'
++₂ > g₂ > max'
mem > null > true > max'
mem > null > false > max'
mem > or > max'
mem > =₂ > max'
mem > not₁ > max'
max > g₂ > max'
max > null > true > max'
max > null > false > max'
max > not₁ > max'

Status:

++₂: [2,1]
=₂: [1,2]
or: []
f₂: [1,2]
true: []
max: []
mem: []
null: []
max': []
g₂: [2,1]
u: []
not₁: [1]
false: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(7) Obligation:

Q DP problem:
P is empty.
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MEM(g(x, y), z) → MEM(x, z)

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MEM(g(x, y), z) → MEM(x, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MEM(x1, x2)  =  MEM(x1, x2)
g(x1, x2)  =  g(x1)
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null(x1)
true  =  true
false  =  false
mem(x1, x2)  =  mem(x1)
or(x1, x2)  =  or(x2)
=(x1, x2)  =  =
max(x1)  =  max(x1)
not(x1)  =  not(x1)
max'(x1, x2)  =  max'
u  =  u

Lexicographic path order with status [LPO].
Precedence:

MEM₂ > nil
f₂ > g₁ > or₁ > nil
f₂ > g₁ > = > nil
++₂ > g₁ > or₁ > nil
++₂ > g₁ > = > nil
mem₁ > null₁ > true > nil
mem₁ > null₁ > false > nil
mem₁ > or₁ > nil
mem₁ > = > nil
mem₁ > not₁ > nil
max₁ > not₁ > nil
max₁ > u > g₁ > or₁ > nil
max₁ > u > g₁ > = > nil
max₁ > u > max' > nil

Status:

null₁: [1]
++₂: [1,2]
=: []
MEM₂: [2,1]
f₂: [2,1]
g₁: [1]
true: []
max': []
u: []
not₁: [1]
or₁: [1]
max₁: [1]
false: []
mem₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(12) Obligation:

Q DP problem:
P is empty.
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

++¹(x, g(y, z)) → ++¹(x, y)

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

++¹(x, g(y, z)) → ++¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2)  =  ++¹(x1, x2)
g(x1, x2)  =  g(x1, x2)
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  x1
true  =  true
false  =  false
mem(x1, x2)  =  x1
or(x1, x2)  =  or(x1, x2)
=(x1, x2)  =  x1
max(x1)  =  max(x1)
not(x1)  =  x1
max'(x1, x2)  =  max'(x1, x2)
u  =  u

Lexicographic path order with status [LPO].
Precedence:

++^1₂ > or₂
f₂ > g₂ > false > or₂
nil > g₂ > false > or₂
nil > true > or₂
++₂ > g₂ > false > or₂
max₁ > g₂ > false > or₂
max₁ > max'₂ > or₂
u > g₂ > false > or₂
u > max'₂ > or₂

Status:

++^1₂: [2,1]
++₂: [1,2]
g₂: [2,1]
max'₂: [1,2]
f₂: [2,1]
u: []
true: []
or₂: [1,2]
max₁: [1]
false: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(17) Obligation:

Q DP problem:
P is empty.
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, g(y, z)) → F(x, y)

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(x, g(y, z)) → F(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
g(x1, x2)  =  g(x1, x2)
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null
true  =  true
false  =  false
mem(x1, x2)  =  mem(x1, x2)
or(x1, x2)  =  or
=(x1, x2)  =  x1
max(x1)  =  max
not(x1)  =  not
max'(x1, x2)  =  max'
u  =  u

Lexicographic path order with status [LPO].
Precedence:

F₂ > g₂
f₂ > nil > g₂
++₂ > g₂
null > true > g₂
null > false > g₂
mem₂ > false > g₂
mem₂ > or > g₂
mem₂ > not > g₂
u > max > max' > g₂

Status:

++₂: [2,1]
or: []
f₂: [2,1]
true: []
not: []
max: []
mem₂: [1,2]
null: []
max': []
g₂: [2,1]
u: []
false: []
F₂: [1,2]
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(22) Obligation:

Q DP problem:
P is empty.
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE