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:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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:

INT(s(x), s(y)) → INT(x, y)
INT(s(x), s(y)) → INTLIST(int(x, y))
INTLIST(cons(x, y)) → INTLIST(y)
INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

INT(s(x), s(y)) → INT(x, y)
INT(s(x), s(y)) → INTLIST(int(x, y))
INTLIST(cons(x, y)) → INTLIST(y)
INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

INT(s(x), s(y)) → INT(x, y)
INTLIST(cons(x, y)) → INTLIST(y)
INT(s(x), s(y)) → INTLIST(int(x, y))
INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

INTLIST(cons(x, y)) → INTLIST(y)

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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.


INTLIST(cons(x, y)) → INTLIST(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INTLIST(x1)  =  INTLIST(x1)
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [2].
Precedence:
cons2 > INTLIST1

The following usable rules [14] were oriented: none



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

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

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

INT(s(x), s(y)) → INT(x, y)
INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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.


INT(s(x), s(y)) → INT(x, y)
The remaining pairs can at least be oriented weakly.

INT(0, s(y)) → INT(s(0), s(y))
Used ordering: Combined order from the following AFS and order.
INT(x1, x2)  =  INT(x2)
s(x1)  =  s(x1)
0  =  0

Recursive Path Order [2].
Precedence:
0 > s1 > INT1

The following usable rules [14] were oriented: none



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

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

INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(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.