Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/SK90SLASH2.34.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.

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

      ↳ QDPAfsSolverProof

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.
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(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)

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(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.