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:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)
INT_LIST(.(x, y)) → INT_LIST(y)
INT(s(x), s(y)) → INT_LIST(int(x, y))

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

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:

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)
INT_LIST(.(x, y)) → INT_LIST(y)
INT(s(x), s(y)) → INT_LIST(int(x, y))

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

INT_LIST(.(x, y)) → INT_LIST(y)

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


INT_LIST(.(x, y)) → INT_LIST(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(INT_LIST(x1)) = (4)x_1   
POL(.(x1, x2)) = 1/4 + (4)x_2   
The value of delta used in the strict ordering is 1.
The following usable rules [17] 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:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

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:

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

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(INT(x1, x2)) = (1/4)x_2   
POL(s(x1)) = 2 + (4)x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.