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))


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
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
Used argument filtering: QUOT3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
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 contains 0 SCCs with 1 less node.