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
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

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:

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

The set Q consists of the following terms:

not1(true)
not1(false)
evenodd2(x0, 0)
evenodd2(0, s1(0))
evenodd2(s1(x0), s1(0))


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

EVENODD2(x, 0) -> EVENODD2(x, s1(0))
EVENODD2(s1(x), s1(0)) -> EVENODD2(x, 0)
EVENODD2(x, 0) -> NOT1(evenodd2(x, s1(0)))

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

The set Q consists of the following terms:

not1(true)
not1(false)
evenodd2(x0, 0)
evenodd2(0, s1(0))
evenodd2(s1(x0), s1(0))

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:

EVENODD2(x, 0) -> EVENODD2(x, s1(0))
EVENODD2(s1(x), s1(0)) -> EVENODD2(x, 0)
EVENODD2(x, 0) -> NOT1(evenodd2(x, s1(0)))

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

The set Q consists of the following terms:

not1(true)
not1(false)
evenodd2(x0, 0)
evenodd2(0, s1(0))
evenodd2(s1(x0), s1(0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPAfsSolverProof

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

EVENODD2(x, 0) -> EVENODD2(x, s1(0))
EVENODD2(s1(x), s1(0)) -> EVENODD2(x, 0)

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

The set Q consists of the following terms:

not1(true)
not1(false)
evenodd2(x0, 0)
evenodd2(0, s1(0))
evenodd2(s1(x0), s1(0))

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.

EVENODD2(s1(x), s1(0)) -> EVENODD2(x, 0)
Used argument filtering: EVENODD2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
QDP
                  ↳ DependencyGraphProof

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

EVENODD2(x, 0) -> EVENODD2(x, s1(0))

The TRS R consists of the following rules:

not1(true) -> false
not1(false) -> true
evenodd2(x, 0) -> not1(evenodd2(x, s1(0)))
evenodd2(0, s1(0)) -> false
evenodd2(s1(x), s1(0)) -> evenodd2(x, 0)

The set Q consists of the following terms:

not1(true)
not1(false)
evenodd2(x0, 0)
evenodd2(0, s1(0))
evenodd2(s1(x0), s1(0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.