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
  ↳ Non-Overlap Check

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.

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:

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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))


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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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:

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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))

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

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

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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.

LE2(s1(X), s1(Y)) -> LE2(X, Y)
Used argument filtering: LE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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
              ↳ 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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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.

IFMINUS3(false, s1(X), Y) -> MINUS2(X, Y)
Used argument filtering: IFMINUS3(x1, x2, x3)  =  x2
s1(x1)  =  s1(x1)
MINUS2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial


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

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

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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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.

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
0  =  0
ifMinus3(x1, x2, x3)  =  x2
Used ordering: Quasi Precedence: s_1 > 0


↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

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)))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
ifMinus3(true, s1(x0), x1)
ifMinus3(false, s1(x0), x1)
quot2(0, s1(x0))
quot2(s1(x0), 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.