Termination w.r.t. Q proof of /tmp/tmpxs-cpb/2.34.trs

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

  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

if(false, x, y) → y

Used ordering:
Polynomial interpretation [25]:

POL(false) = 1
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(true) = 0
POL(u) = 0
POL(v) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

if(true, x, y) → x
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.

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

if(true, x, y) → x
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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

if(true, x, y) → x

Used ordering:
Polynomial interpretation [25]:

POL(if(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + 2·x₃
POL(true) = 1
POL(u) = 0
POL(v) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

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.

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

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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

if(x, y, y) → y

Used ordering:
Polynomial interpretation [25]:

POL(if(x₁, x₂, x₃)) = 2 + 2·x₁ + x₂ + x₃
POL(u) = 0
POL(v) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

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

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(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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(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 approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

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)

The TRS R consists of the following rules:

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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

IF(if(x, y, z), u, v) → IF(z, u, v)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
IF(if(x, y, z), u, v) → IF(y, u, v)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3