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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 UsableRulesProof (⇔)
↳10 QDP
↳11 QDPSizeChangeProof (⇔)
↳12 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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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


(2) Obligation:

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.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(if(x1, x2, x3)) = 2 + 2·x1 + x2 + x3   
POL(u) = 0   
POL(v) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if(x, y, y) → y


(4) Obligation:

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.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) 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(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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

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.

(9) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

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.

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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

(12) TRUE