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:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.

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

*12(p1(x), y) -> *12(x, y)
MINUS1(p1(x)) -> MINUS1(x)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> P1(+2(x, y))
+12(s1(x), y) -> S1(+2(x, y))
+12(p1(x), y) -> +12(x, y)
MINUS1(p1(x)) -> S1(minus1(x))
*12(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
MINUS1(s1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> P1(minus1(x))
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(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:

*12(p1(x), y) -> *12(x, y)
MINUS1(p1(x)) -> MINUS1(x)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> P1(+2(x, y))
+12(s1(x), y) -> S1(+2(x, y))
+12(p1(x), y) -> +12(x, y)
MINUS1(p1(x)) -> S1(minus1(x))
*12(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
MINUS1(s1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> P1(minus1(x))
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

MINUS1(s1(x)) -> MINUS1(x)
MINUS1(p1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

MINUS1(p1(x)) -> MINUS1(x)
Used argument filtering: MINUS1(x1)  =  x1
s1(x1)  =  x1
p1(x1)  =  p1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

MINUS1(s1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

MINUS1(s1(x)) -> MINUS1(x)
Used argument filtering: MINUS1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

+12(p1(x), y) -> +12(x, y)
Used argument filtering: +12(x1, x2)  =  x1
s1(x1)  =  x1
p1(x1)  =  p1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

+12(s1(x), y) -> +12(x, y)
Used argument filtering: +12(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

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

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(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
          ↳ QDP
            ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

*12(p1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x1
s1(x1)  =  x1
p1(x1)  =  p1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
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.

*12(s1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

p1(s1(x)) -> x
s1(p1(x)) -> x
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(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.