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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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:

QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → PLUS(z, s(0))
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(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:

QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → PLUS(z, s(0))
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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.


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

POL(PLUS(x1, x2)) = (2)x_1   
POL(s(x1)) = 4 + (4)x_1   
The value of delta used in the strict ordering is 8.
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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(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:

QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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.


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

QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
Used ordering: Polynomial interpretation [25,35]:

POL(plus(x1, x2)) = 0   
POL(QUOT(x1, x2, x3)) = (4)x_1   
POL(s(x1)) = 4 + (2)x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



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

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

QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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.


QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(plus(x1, x2)) = (4)x_2   
POL(QUOT(x1, x2, x3)) = (2)x_2   
POL(s(x1)) = 0   
POL(0) = 4   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

plus(0, y) → y
plus(s(x), y) → s(plus(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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(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.