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:

if(true, x, y) → x
if(false, x, y) → y
if(x, y, y) → y
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(x, y, z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

if(true, x, y) → x
if(false, x, y) → y
if(x, y, y) → y
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(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:

IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)

The TRS R consists of the following rules:

if(true, x, y) → x
if(false, x, y) → y
if(x, y, y) → y
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(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:

IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)

The TRS R consists of the following rules:

if(true, x, y) → x
if(false, x, y) → y
if(x, y, y) → y
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(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.


IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  IF(x1, x2, x3)
if(x1, x2, x3)  =  if(x1, x2, x3)
true  =  true
false  =  false

Recursive path order with status [2].
Precedence:
IF3 > if3
true > if3
false > if3

Status:
true: multiset
if3: [1,3,2]
false: multiset
IF3: [1,2,3]

The following usable rules [14] were oriented:

if(true, x, y) → x
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(x, y, z)
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, y, y) → y
if(false, x, y) → y



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

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

if(true, x, y) → x
if(false, x, y) → y
if(x, y, y) → y
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) → if(x, y, z)
if(x, y, if(x, y, z)) → if(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.