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))
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.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

true > if₃
false > if₃

Status:

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

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE