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)
quot3(x, 0, s1(z)) -> s1(quot3(x, s1(z), 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)
quot3(x, 0, s1(z)) -> s1(quot3(x, s1(z), 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:

QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> QUOT3(x, s1(z), 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)
quot3(x, 0, s1(z)) -> s1(quot3(x, s1(z), s1(z)))

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

↳ QTRS
  ↳ DependencyPairsProof
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, s1(z), 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)
quot3(x, 0, s1(z)) -> s1(quot3(x, s1(z), 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, s1(z), s1(z))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUOT3(x1, x2, x3)) = 3·x1 + x1·x3   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

QUOT3(x, 0, s1(z)) -> QUOT3(x, s1(z), 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)
quot3(x, 0, s1(z)) -> s1(quot3(x, s1(z), 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 0 SCCs with 1 less node.