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:

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

Q is empty.

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

MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> OR2(mem2(x, y), mem2(x, z))
MEM2(x, union2(y, z)) -> MEM2(x, y)

The TRS R consists of the following rules:

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

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:

MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> OR2(mem2(x, y), mem2(x, z))
MEM2(x, union2(y, z)) -> MEM2(x, y)

The TRS R consists of the following rules:

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

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

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

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

MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> MEM2(x, y)

The TRS R consists of the following rules:

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

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.

MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> MEM2(x, y)
Used argument filtering: MEM2(x1, x2)  =  x2
union2(x1, x2)  =  union2(x1, x2)
Used ordering: Quasi Precedence: trivial 


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

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

or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))

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.