Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MSORT(.(x, y)) -> MIN(x, y)
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
MSORT(.(x, y)) -> DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) -> MIN(x, z)
MIN(x, .(y, z)) -> MIN(y, z)
DEL(x, .(y, z)) -> DEL(x, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x, .(y, z)) -> MIN(y, z)
MIN(x, .(y, z)) -> MIN(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MIN(x, .(y, z)) -> MIN(x, z)

one new Dependency Pair is created:

MIN(x'', .(y, .(y'', z''))) -> MIN(x'', .(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x'', .(y, .(y'', z''))) -> MIN(x'', .(y'', z''))
MIN(x, .(y, z)) -> MIN(y, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MIN(x, .(y, z)) -> MIN(y, z)

two new Dependency Pairs are created:

MIN(x, .(y0, .(y'', z''))) -> MIN(y0, .(y'', z''))
MIN(x, .(y0, .(y'', .(y'''', z'''')))) -> MIN(y0, .(y'', .(y'''', z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x, .(y0, .(y'', .(y'''', z'''')))) -> MIN(y0, .(y'', .(y'''', z'''')))
MIN(x, .(y0, .(y'', z''))) -> MIN(y0, .(y'', z''))
MIN(x'', .(y, .(y'', z''))) -> MIN(x'', .(y'', z''))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MIN(x'', .(y, .(y'', z''))) -> MIN(x'', .(y'', z''))

three new Dependency Pairs are created:

MIN(x'''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x'''', .(y''0, .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x''', .(y''0, .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(x''', .(y''0, .(y'''', .(y'''''', z''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x''', .(y, .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(x''', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x''', .(y''0, .(y'''', z'''')))
MIN(x'''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x'''', .(y''0, .(y'''', z'''')))
MIN(x, .(y0, .(y'', z''))) -> MIN(y0, .(y'', z''))
MIN(x, .(y0, .(y'', .(y'''', z'''')))) -> MIN(y0, .(y'', .(y'''', z'''')))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MIN(x, .(y0, .(y'', z''))) -> MIN(y0, .(y'', z''))

four new Dependency Pairs are created:

MIN(x, .(y0'', .(y''0, .(y'''', z'''')))) -> MIN(y0'', .(y''0, .(y'''', z'''')))
MIN(x, .(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', z'''')))) -> MIN(y0'', .(y''0, .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x''', .(y''0, .(y'''', z'''')))
MIN(x'''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x'''', .(y''0, .(y'''', z'''')))
MIN(x, .(y0, .(y'', .(y'''', z'''')))) -> MIN(y0, .(y'', .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(x''', .(y''0, .(y'''', .(y'''''', z''''''))))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MIN(x, .(y0, .(y'', .(y'''', z'''')))) -> MIN(y0, .(y'', .(y'''', z'''')))

five new Dependency Pairs are created:

MIN(x, .(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pairs:

MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))
MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', z'''')))) -> MIN(y0'', .(y''0, .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(x''', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x''', .(y''0, .(y'''', z'''')))
MIN(x'''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x'''', .(y''0, .(y'''', z'''')))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', .(y'''''''''', z''''''''''))))))
MIN(x, .(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y'''''', .(y''0'''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', .(y'''''''', z'''''''')))))
MIN(x, .(y0', .(y''', .(y''''0, .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y''''0, .(y'''''', z''''''))))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', z''''''))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(y0'', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x, .(y0'', .(y''0, .(y'''', z'''')))) -> MIN(y0'', .(y''0, .(y'''', z'''')))
MIN(x''', .(y, .(y''0, .(y'''', .(y'''''', z''''''))))) -> MIN(x''', .(y''0, .(y'''', .(y'''''', z''''''))))
MIN(x''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x''', .(y''0, .(y'''', z'''')))
MIN(x'''', .(y, .(y''0, .(y'''', z'''')))) -> MIN(x'''', .(y''0, .(y'''', z'''')))
MIN(x, .(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))) -> MIN(y0', .(y''', .(y''0'', .(y'''''', .(y'''''''', z'''''''')))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MIN(x₁, x₂)) = 1 + x₂

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Rw

Dependency Pair:

DEL(x, .(y, z)) -> DEL(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

DEL(x, .(y, z)) -> DEL(x, z)

one new Dependency Pair is created:

DEL(x'', .(y, .(y'', z''))) -> DEL(x'', .(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Rw

Dependency Pair:

DEL(x'', .(y, .(y'', z''))) -> DEL(x'', .(y'', z''))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

DEL(x'', .(y, .(y'', z''))) -> DEL(x'', .(y'', z''))

one new Dependency Pair is created:

DEL(x'''', .(y, .(y''0, .(y'''', z'''')))) -> DEL(x'''', .(y''0, .(y'''', z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Rw

Dependency Pair:

DEL(x'''', .(y, .(y''0, .(y'''', z'''')))) -> DEL(x'''', .(y''0, .(y'''', z'''')))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DEL(x'''', .(y, .(y''0, .(y'''', z'''')))) -> DEL(x'''', .(y''0, .(y'''', z'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(DEL(x₁, x₂)) = 1 + x₂

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rewriting Transformation

Dependency Pair:

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

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

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

one new Dependency Pair is created:

MSORT(.(x, y)) -> MSORT(if(=(min(x, y), x), y, .(x, del(min(x, y), y))))

The transformation is resulting in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes

POL(MIN(x₁, x₂))	= 1 + x₂
POL(.(x₁, x₂))	= 1 + x₂

POL(DEL(x₁, x₂))	= 1 + x₂
POL(.(x₁, x₂))	= 1 + x₂