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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 PisEmptyProof (⇔)
↳6 TRUE

(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: Recursive path order with status [RPO].
Quasi-Precedence:
IF3 > [if3, u, v]
true > [if3, u, v]
false > [if3, u, v]

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


The following usable rules [FROCOS05] were oriented:

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

(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