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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE

(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)  =  x1
or(x1, x2)  =  x2
=(x1, x2)  =  x1
max(x1)  =  max(x1)
not(x1)  =  not
max'(x1, x2)  =  max'(x2)

Recursive Path Order [RPO].
Precedence:
f2 > g2 > false > not
f2 > g2 > max'1 > not
f2 > nil > false > not
f2 > nil > max'1 > not
++2 > g2 > false > not
++2 > g2 > max'1 > not
max1 > u > g2 > false > not
max1 > u > g2 > max'1 > not
max1 > null > true > not
max1 > null > false > not

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)  =  x1
g(x1, x2)  =  g(x1)
f(x1, x2)  =  f(x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null
true  =  true
false  =  false
mem(x1, x2)  =  mem(x1)
or(x1, x2)  =  or
=(x1, x2)  =  x2
max(x1)  =  max(x1)
not(x1)  =  x1
max'(x1, x2)  =  max'
u  =  u

Recursive Path Order [RPO].
Precedence:
f1 > g1 > u > false
f1 > nil > false
++2 > g1 > u > false
mem1 > null > true > false
mem1 > or > false
max1 > g1 > u > false
max1 > null > true > false
max1 > max' > false

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)  =  x2
g(x1, x2)  =  g(x1)
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null
true  =  true
false  =  false
mem(x1, x2)  =  mem(x1)
or(x1, x2)  =  or
=(x1, x2)  =  x2
max(x1)  =  max(x1)
not(x1)  =  x1
max'(x1, x2)  =  max'
u  =  u

Recursive Path Order [RPO].
Precedence:
f2 > g1 > false
f2 > g1 > u
f2 > nil > false
++2 > g1 > false
++2 > g1 > u
mem1 > null > true
mem1 > null > false
mem1 > or
max1 > g1 > false
max1 > g1 > u
max1 > null > true
max1 > null > false
max1 > max'

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)  =  x2
g(x1, x2)  =  g(x1)
f(x1, x2)  =  f(x1, x2)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
null(x1)  =  null
true  =  true
false  =  false
mem(x1, x2)  =  mem(x1)
or(x1, x2)  =  or
=(x1, x2)  =  x2
max(x1)  =  max(x1)
not(x1)  =  x1
max'(x1, x2)  =  max'
u  =  u

Recursive Path Order [RPO].
Precedence:
f2 > g1 > false
f2 > g1 > u
f2 > nil > false
++2 > g1 > false
++2 > g1 > u
mem1 > null > true
mem1 > null > false
mem1 > or
max1 > g1 > false
max1 > g1 > u
max1 > null > true
max1 > null > false
max1 > max'

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