Term Rewriting System R:
[x, y, z]
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))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

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))


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))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))
six new Dependency Pairs are created:

IF(if(x, true, z), u, v) -> IF(x, u, if(z, u, v))
IF(if(x, false, z), u, v) -> IF(x, v, if(z, u, v))
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, y, true), u, v) -> IF(x, if(y, u, v), u)
IF(if(x, y, false), u, v) -> IF(x, if(y, u, v), v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Forward Instantiation Transformation


Dependency Pairs:

IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))
IF(if(x, y, false), u, v) -> IF(x, if(y, u, v), v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, true, z), u, v) -> IF(x, u, 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)


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))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF(if(x, y, z), u, v) -> IF(y, u, v)
five new Dependency Pairs are created:

IF(if(x, if(x'', y'', z''), z), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, if(x'', true, z''), z), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', if(x'''', y'''', z''''), z''), z), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, if(x'', y'', false), z), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, if(x'', y'', if(x'''', y'''', z'''')), z), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 3
Forward Instantiation Transformation


Dependency Pairs:

IF(if(x, if(x'', y'', if(x'''', y'''', z'''')), z), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)
IF(if(x, if(x'', y'', false), z), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, if(x'', if(x'''', y'''', z''''), z''), z), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, if(x'', true, z''), z), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, y, false), u, v) -> IF(x, if(y, u, v), v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, true, z), u, v) -> IF(x, u, if(z, u, v))
IF(if(x, y, z), u, v) -> IF(z, u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))


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))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF(if(x, y, z), u, v) -> IF(z, u, v)
nine new Dependency Pairs are created:

IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, y, if(x'', true, z'')), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', z''''), z'')), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', y'', false)), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, y, if(x'', y'', if(x'''', y'''', z''''))), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)
IF(if(x, y, if(x'', if(x'''', true, z''''), z'')), u, v) -> IF(if(x'', if(x'''', true, z''''), z''), u, v)
IF(if(x, y, if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z'')), u, v) -> IF(if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', false), z'')), u, v) -> IF(if(x'', if(x'''', y'''', false), z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z'')), u, v) -> IF(if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z''), u, v)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 4
Polynomial Ordering


Dependency Pairs:

IF(if(x, y, if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z'')), u, v) -> IF(if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', false), z'')), u, v) -> IF(if(x'', if(x'''', y'''', false), z''), u, v)
IF(if(x, y, if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z'')), u, v) -> IF(if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z''), u, v)
IF(if(x, y, if(x'', if(x'''', true, z''''), z'')), u, v) -> IF(if(x'', if(x'''', true, z''''), z''), u, v)
IF(if(x, y, if(x'', y'', if(x'''', y'''', z''''))), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)
IF(if(x, y, if(x'', y'', false)), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', z''''), z'')), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', true, z'')), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', false), z), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, if(x'', if(x'''', y'''', z''''), z''), z), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, if(x'', true, z''), z), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))
IF(if(x, y, false), u, v) -> IF(x, if(y, u, v), v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, true, z), u, v) -> IF(x, u, if(z, u, v))
IF(if(x, if(x'', y'', if(x'''', y'''', z'''')), z), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(if(x, y, false), u, v) -> IF(x, if(y, u, v), v)


Additionally, the following usable rules for innermost can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(v)=  0  
  POL(false)=  1  
  POL(true)=  0  
  POL(u)=  0  
  POL(IF(x1, x2, x3))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 5
Polynomial Ordering


Dependency Pairs:

IF(if(x, y, if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z'')), u, v) -> IF(if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', false), z'')), u, v) -> IF(if(x'', if(x'''', y'''', false), z''), u, v)
IF(if(x, y, if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z'')), u, v) -> IF(if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z''), u, v)
IF(if(x, y, if(x'', if(x'''', true, z''''), z'')), u, v) -> IF(if(x'', if(x'''', true, z''''), z''), u, v)
IF(if(x, y, if(x'', y'', if(x'''', y'''', z''''))), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)
IF(if(x, y, if(x'', y'', false)), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', z''''), z'')), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', true, z'')), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', false), z), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, if(x'', if(x'''', y'''', z''''), z''), z), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, if(x'', true, z''), z), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, true, z), u, v) -> IF(x, u, if(z, u, v))
IF(if(x, if(x'', y'', if(x'''', y'''', z'''')), z), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(if(x, true, z), u, v) -> IF(x, u, if(z, u, v))


Additionally, the following usable rules for innermost can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(v)=  0  
  POL(false)=  0  
  POL(true)=  1  
  POL(u)=  0  
  POL(IF(x1, x2, x3))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IF(if(x, y, if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z'')), u, v) -> IF(if(x'', if(x'''', y'''', if(x'''''', y'''''', z'''''')), z''), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', false), z'')), u, v) -> IF(if(x'', if(x'''', y'''', false), z''), u, v)
IF(if(x, y, if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z'')), u, v) -> IF(if(x'', if(x'''', if(x'''''', y'''''', z''''''), z''''), z''), u, v)
IF(if(x, y, if(x'', if(x'''', true, z''''), z'')), u, v) -> IF(if(x'', if(x'''', true, z''''), z''), u, v)
IF(if(x, y, if(x'', y'', if(x'''', y'''', z''''))), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)
IF(if(x, y, if(x'', y'', false)), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, y, if(x'', if(x'''', y'''', z''''), z'')), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', true, z'')), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', false), z), u, v) -> IF(if(x'', y'', false), u, v)
IF(if(x, if(x'', if(x'''', y'''', z''''), z''), z), u, v) -> IF(if(x'', if(x'''', y'''', z''''), z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, if(x'', true, z''), z), u, v) -> IF(if(x'', true, z''), u, v)
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(if(x'', y'', z''), u, v)
IF(if(x, y, if(x'', y'', z'')), u, v) -> IF(x, if(y, u, v), if(x'', if(y'', u, v), if(z'', u, v)))
IF(if(x, if(x'', y'', z''), z), u, v) -> IF(x, if(x'', if(y'', u, v), if(z'', u, v)), if(z, u, v))
IF(if(x, if(x'', y'', if(x'''', y'''', z'''')), z), u, v) -> IF(if(x'', y'', if(x'''', y'''', z'''')), u, v)


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))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:10 minutes