Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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)
IF(if(x, y, z), u, v) → IF(z, 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)
IF(if(x, y, z), u, v) → IF(z, u, v)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  IF(x1)
if(x1, x2, x3)  =  if(x1, x2, x3)
u  =  u
v  =  v
true  =  true
false  =  false

Lexicographic path order with status [LPO].
Precedence:

u > IF₁ > if₃
u > v > if₃
true > if₃
false > if₃

Status:

if₃: [1,2,3]
v: []
u: []
true: []
false: []
IF₁: [1]

The following usable rules [FROCOS05] were oriented:

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))

(4) Obligation:

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE