Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.

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

MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
MEM2(g2(x, y), z) -> MEM2(x, z)
F2(x, g2(y, z)) -> F2(x, y)
++12(x, g2(y, z)) -> ++12(x, y)
MEM2(x, max1(x)) -> NULL1(x)

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
MEM2(g2(x, y), z) -> MEM2(x, z)
F2(x, g2(y, z)) -> F2(x, y)
++12(x, g2(y, z)) -> ++12(x, y)
MEM2(x, max1(x)) -> NULL1(x)

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
Used argument filtering: MAX1(x1)  =  x1
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

MEM2(g2(x, y), z) -> MEM2(x, z)

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MEM2(g2(x, y), z) -> MEM2(x, z)
Used argument filtering: MEM2(x1, x2)  =  x1
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

++12(x, g2(y, z)) -> ++12(x, y)

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++12(x, g2(y, z)) -> ++12(x, y)
Used argument filtering: ++12(x1, x2)  =  x2
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof

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

F2(x, g2(y, z)) -> F2(x, y)

The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F2(x, g2(y, z)) -> F2(x, y)
Used argument filtering: F2(x1, x2)  =  x2
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.