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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 4 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( g2(x1, x2) ) = x1 + x2 + 1


POL( u ) = 1


POL( MAX1(x1) ) = x1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MEM2(g2(x, y), z) -> MEM2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( g2(x1, x2) ) = x1 + 1


POL( MEM2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


++12(x, g2(y, z)) -> ++12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ++12(x1, x2) ) = x2


POL( g2(x1, x2) ) = x1 + x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(x, g2(y, z)) -> F2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( g2(x1, x2) ) = x1 + x2 + 1


POL( F2(x1, x2) ) = x2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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.