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:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> PLUS2(z, s1(0))
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> PLUS2(z, s1(0))
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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.


PLUS2(s1(x), y) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

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:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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:

QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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.


QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
The remaining pairs can at least be oriented weakly.

QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUOT3(x1, x2, x3)) = x1   
POL(plus2(x1, x2)) = 0   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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.


QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(QUOT3(x1, x2, x3)) = x2   
POL(plus2(x1, x2)) = x2   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented:

plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))



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

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

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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.