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:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))

Q is empty.

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

ODD1(s1(x)) -> NOT1(odd1(x))
+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
ODD1(s1(x)) -> ODD1(x)

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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:

ODD1(s1(x)) -> NOT1(odd1(x))
+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
ODD1(s1(x)) -> ODD1(x)

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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(x, s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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: Precedence:
trivial



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

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

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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

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

ODD1(s1(x)) -> ODD1(x)

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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.

ODD1(s1(x)) -> ODD1(x)
Used argument filtering: ODD1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



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

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

not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, 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.