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))

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))

Q is empty.

Using Dependency Pairs [1,15] 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(y, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, 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.

↳ 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(y, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, 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.
We use the reduction pair processor [15].


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(y, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(if(x1, x2, x3)) = 1/4 + x_1 + x_2 + (4)x_3   
POL(v) = 0   
POL(u) = 0   
POL(true) = 0   
POL(false) = 0   
POL(IF(x1, x2, x3)) = (1/2)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none



↳ 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))

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.