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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PLUS2(s1(X), Y) -> PLUS2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x1 + 2·x1·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(MIN2(x1, x2)) = x1   
POL(Z) = 0   
POL(min2(x1, x2)) = 2 + 3·x1 + 3·x1·x2 + 3·x2   
POL(plus2(x1, x2)) = 0   
POL(s1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUOT2(x1, x2)) = 2·x1·x2   
POL(Z) = 0   
POL(min2(x1, x2)) = x1   
POL(plus2(x1, x2)) = 0   
POL(s1(x1)) = 3 + 2·x1 + 2·x12   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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.