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:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

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:

-12(x, s1(y)) -> P1(s1(y))
-12(x, s1(y)) -> -12(x, p1(s1(y)))

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

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:

-12(x, s1(y)) -> P1(s1(y))
-12(x, s1(y)) -> -12(x, p1(s1(y)))

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

-12(x, s1(y)) -> -12(x, p1(s1(y)))

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

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.


-12(x, s1(y)) -> -12(x, p1(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( -12(x1, x2) ) = max{0, x1 + 3x2 - 1}


POL( p1(x1) ) = max{0, x1 - 3}



The following usable rules [14] were oriented:

p1(s1(x)) -> x



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

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.