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:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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:

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)
MINUS2(s1(X), Y) -> LE2(s1(X), Y)
MINUS2(s1(X), Y) -> IFMINUS3(le2(s1(X), Y), s1(X), Y)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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:

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)
MINUS2(s1(X), Y) -> LE2(s1(X), Y)
MINUS2(s1(X), Y) -> IFMINUS3(le2(s1(X), Y), s1(X), Y)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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:

LE2(s1(X), s1(Y)) -> LE2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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.


LE2(s1(X), s1(Y)) -> LE2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 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:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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:

IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)
MINUS2(s1(X), Y) -> IFMINUS3(le2(s1(X), Y), s1(X), Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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.


MINUS2(s1(X), Y) -> IFMINUS3(le2(s1(X), Y), s1(X), Y)
The remaining pairs can at least be oriented weakly.

IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(IFMINUS3(x1, x2, x3)) = x2 + 2·x3   
POL(MINUS2(x1, x2)) = 3 + 3·x1 + 2·x2   
POL(false) = 0   
POL(le2(x1, x2)) = 0   
POL(s1(x1)) = 3 + 3·x1   
POL(true) = 0   

The following usable rules [14] were oriented: none



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

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

IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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 0 SCCs with 1 less node.

↳ 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(minus2(X, Y), s1(Y))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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(minus2(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   
POL(false) = 0   
POL(ifMinus3(x1, x2, x3)) = x2   
POL(le2(x1, x2)) = 0   
POL(minus2(x1, x2)) = x1   
POL(s1(x1)) = 2 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
minus2(0, Y) -> 0
ifMinus3(true, s1(X), Y) -> 0



↳ 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:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
minus2(0, Y) -> 0
minus2(s1(X), Y) -> ifMinus3(le2(s1(X), Y), s1(X), Y)
ifMinus3(true, s1(X), Y) -> 0
ifMinus3(false, s1(X), Y) -> s1(minus2(X, Y))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(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.