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:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.

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

PLUS2(s1(X), Y) -> PLUS2(X, Y)
MIN2(min2(X, Y), Z) -> PLUS2(Y, Z)
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))

The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

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:

PLUS2(s1(X), Y) -> PLUS2(X, Y)
MIN2(min2(X, Y), Z) -> PLUS2(Y, Z)
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))

The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ 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:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))

The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.
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.

MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
Used argument filtering: MIN2(x1, x2)  =  x1
s1(x1)  =  x1
min2(x1, x2)  =  min1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)

The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.
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.

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
Used argument filtering: MIN2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

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
          ↳ QDP
            ↳ QDPAfsSolverProof

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

QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))

The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

Q is empty.
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.

QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
min2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))

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.