Termination w.r.t. Q proof of TRS/SK902.34.trs

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

if₃(true, x, y) -> x
if₃(false, x, y) -> y
if₃(x, y, y) -> y
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

      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

if₃(true, x, y) -> x
if₃(false, x, y) -> y
if₃(x, y, y) -> y
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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( v ) = 0

POL( true ) = 3

POL( if₃(x₁, ..., x₃) ) = x₁ + x₂ + x₃ + 1

POL( u ) = 1

POL( false ) = 0

POL( IF₃(x₁, ..., x₃) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

if₃(true, x, y) -> x
if₃(false, x, y) -> y
if₃(x, y, y) -> y
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.