Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/SK90SLASH4.23.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))
if₃(x, if₃(x, y, z), z) -> if₃(x, y, z)
if₃(x, y, if₃(x, y, z)) -> if₃(x, y, z)

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

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

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

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 strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
IF₃(x1, x2, x3) = IF₃(x1, x2, x3)
if₃(x1, x2, x3) = if₃(x1, x2, x3)
true = true
false = false

Lexicographic Path Order [19].
Precedence:

true > [IF₃, if_3]
false > [IF₃, if_3]

The following usable rules [14] were oriented:

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

↳ 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))
if₃(x, if₃(x, y, z), z) -> if₃(x, y, z)
if₃(x, y, if₃(x, y, z)) -> if₃(x, y, 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.