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:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

INT_LIST1(.2(x, y)) -> INT_LIST1(y)
INT2(s1(x), s1(y)) -> INT2(x, y)
INT2(s1(x), s1(y)) -> INT_LIST1(int2(x, y))
INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

INT_LIST1(.2(x, y)) -> INT_LIST1(y)
INT2(s1(x), s1(y)) -> INT2(x, y)
INT2(s1(x), s1(y)) -> INT_LIST1(int2(x, y))
INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

INT_LIST1(.2(x, y)) -> INT_LIST1(y)

The TRS R consists of the following rules:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


INT_LIST1(.2(x, y)) -> INT_LIST1(y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
INT_LIST1(x1)  =  INT_LIST1(x1)
.2(x1, x2)  =  .2(x1, x2)

Lexicographic Path Order [19].
Precedence:
.2 > INTLIST1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

INT2(s1(x), s1(y)) -> INT2(x, y)
INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


INT2(s1(x), s1(y)) -> INT2(x, y)
The remaining pairs can at least by weakly be oriented.

INT2(0, s1(y)) -> INT2(s1(0), s1(y))
Used ordering: Combined order from the following AFS and order.
INT2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
0 > s1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ DependencyGraphProof

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

INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

int2(0, 0) -> .2(0, nil)
int2(0, s1(y)) -> .2(0, int2(s1(0), s1(y)))
int2(s1(x), 0) -> nil
int2(s1(x), s1(y)) -> int_list1(int2(x, y))
int_list1(nil) -> nil
int_list1(.2(x, y)) -> .2(s1(x), int_list1(y))

The set Q consists of the following terms:

int2(0, 0)
int2(0, s1(x0))
int2(s1(x0), 0)
int2(s1(x0), s1(x1))
int_list1(nil)
int_list1(.2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.