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 oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(IF3(x1, x2, x3)) = 2·x1   
POL(false) = 0   
POL(if3(x1, x2, x3)) = 1 + 2·x1 + 2·x2 + 2·x3   
POL(true) = 0   

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))
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.