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:

if3(if3(x, y, z), u, v) -> if3(x, if3(y, u, v), if3(z, u, v))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

if3(if3(x, y, z), u, v) -> if3(x, if3(y, u, v), if3(z, u, v))

Q is empty.

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

IF3(if3(x, y, z), u, v) -> IF3(z, u, v)
IF3(if3(x, y, z), u, v) -> IF3(x, if3(y, u, v), if3(z, u, v))
IF3(if3(x, y, z), u, v) -> IF3(y, u, v)

The TRS R consists of the following rules:

if3(if3(x, y, z), u, v) -> if3(x, if3(y, u, v), if3(z, u, v))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

IF3(if3(x, y, z), u, v) -> IF3(z, u, v)
IF3(if3(x, y, z), u, v) -> IF3(x, if3(y, u, v), if3(z, u, v))
IF3(if3(x, y, z), u, v) -> IF3(y, u, v)

The TRS R consists of the following rules:

if3(if3(x, y, z), u, v) -> if3(x, if3(y, u, v), if3(z, u, v))

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.

IF3(if3(x, y, z), u, v) -> IF3(z, u, v)
IF3(if3(x, y, z), u, v) -> IF3(x, if3(y, u, v), if3(z, u, v))
IF3(if3(x, y, z), u, v) -> IF3(y, u, v)
Used argument filtering: IF3(x1, x2, x3)  =  x1
if3(x1, x2, x3)  =  if3(x1, x2, x3)
Used ordering: Quasi Precedence: trivial 


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

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

if3(if3(x, y, z), u, v) -> if3(x, if3(y, u, v), if3(z, u, v))

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.