Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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.

     ↳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.

     ↳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:
0:12 minutes

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