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
  ↳ 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)))

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:

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)))

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:

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)))

Q is empty.
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
  ↳ 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)))

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), s1(Y)) -> MINUS2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ 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)))

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

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)))

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.


DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 2x1 + 3


POL( 0 ) = 1


POL( minus2(x1, x2) ) = x1 + 1


POL( DIV2(x1, x2) ) = 2x1 + 2x2


POL( p1(x1) ) = max{0, 2x1 - 1}



The following usable rules [14] were oriented:

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



↳ 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)))

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.