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
  ↳ 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)
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.

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)
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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))


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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
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.

PLUS2(s1(x), y) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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, 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
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)
plus2(x1, x2)  =  plus2(x1, x2)
0  =  0
Used ordering: Quasi Precedence: plus_2 > s_1 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof

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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
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(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))
Used argument filtering: QUOT3(x1, x2, x3)  =  x2
0  =  0
plus2(x1, x2)  =  x2
s1(x1)  =  s
Used ordering: Quasi Precedence: 0 > s 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ 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)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.