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:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

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:

INT2(s1(x), s1(y)) -> INT2(x, y)
INT2(s1(x), s1(y)) -> INTLIST1(int2(x, y))
INTLIST1(cons2(x, y)) -> INTLIST1(y)
INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

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:

INT2(s1(x), s1(y)) -> INT2(x, y)
INT2(s1(x), s1(y)) -> INTLIST1(int2(x, y))
INTLIST1(cons2(x, y)) -> INTLIST1(y)
INT2(0, s1(y)) -> INT2(s1(0), s1(y))

The TRS R consists of the following rules:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

INTLIST1(cons2(x, y)) -> INTLIST1(y)

The TRS R consists of the following rules:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

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.


INTLIST1(cons2(x, y)) -> INTLIST1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( INTLIST1(x1) ) = max{0, x1 - 2}


POL( cons2(x1, x2) ) = x2 + 3



The following usable rules [14] were oriented: none



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

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

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

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

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:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

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.


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

INT2(0, s1(y)) -> INT2(s1(0), s1(y))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( INT2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ 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:

intlist1(nil) -> nil
int2(s1(x), 0) -> nil
int2(x, x) -> cons2(x, nil)
intlist1(cons2(x, y)) -> cons2(s1(x), intlist1(y))
int2(s1(x), s1(y)) -> intlist1(int2(x, y))
int2(0, s1(y)) -> cons2(0, int2(s1(0), s1(y)))
intlist1(cons2(x, nil)) -> cons2(s1(x), nil)

Q is empty.
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.