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:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(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:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(s1(x0), s1(x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MINUS2(s1(X), s1(Y)) -> P1(minus2(X, Y))
DIV2(s1(X), s1(Y)) -> MINUS2(X, Y)
DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(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:

MINUS2(s1(X), s1(Y)) -> P1(minus2(X, Y))
DIV2(s1(X), s1(Y)) -> MINUS2(X, Y)
DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(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
                ↳ QDPOrderProof

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

DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))

The TRS R consists of the following rules:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DIV2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
p1(x1)  =  p1(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
[s1, p1]


The following usable rules [14] were oriented:

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

minus2(X, 0) -> X
minus2(s1(X), s1(Y)) -> p1(minus2(X, Y))
p1(s1(X)) -> X
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> s1(div2(minus2(X, Y), s1(Y)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
p1(s1(x0))
div2(0, s1(x0))
div2(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.