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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
quot3(x0, 0, s1(x1))


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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
quot3(x0, 0, s1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
quot3(x0, 0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

QUOT3(x, 0, s1(z)) -> QUOT3(x, s1(z), s1(z))
Used ordering: Combined order from the following AFS and order.
QUOT3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
quot3(x0, 0, s1(x1))

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.