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)))

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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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)))

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

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)))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

= > nil
if > nil
{msort, .} > MIN > nil
{msort, .} > {min, <=} > nil
del > nil

resulting in one new DP problem.
Used Argument Filtering System:

MIN(x₁, x₂) -> MIN(x₁, x₂)
.(x₁, x₂) -> .(x₁, x₂)
msort(x₁) -> msort(x₁)
min(x₁, x₂) -> min(x₁, x₂)
del(x₁, x₂) -> x₁
if(x₁, x₂, x₃) -> x₁
<=(x₁, x₂) -> <=(x₁, x₂)
=(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

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)))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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)))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

= > nil
if > nil
msort > . > min > <= > nil
DEL > nil
del > nil

resulting in one new DP problem.
Used Argument Filtering System:

DEL(x₁, x₂) -> DEL(x₁, x₂)
.(x₁, x₂) -> .(x₁, x₂)
msort(x₁) -> msort(x₁)
min(x₁, x₂) -> min(x₁, x₂)
del(x₁, x₂) -> x₁
if(x₁, x₂, x₃) -> x₁
<=(x₁, x₂) -> <=(x₁, x₂)
=(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

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)))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

= > nil
if > nil
msort > . > {min, <=} > nil
MSORT > nil
del > nil

resulting in one new DP problem.
Used Argument Filtering System:

MSORT(x₁) -> MSORT(x₁)
.(x₁, x₂) -> .(x₁, x₂)
del(x₁, x₂) -> x₁
min(x₁, x₂) -> min(x₁, x₂)
if(x₁, x₂, x₃) -> x₁
=(x₁, x₂) -> x₁
<=(x₁, x₂) -> <=(x₁, x₂)
msort(x₁) -> msort(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

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)))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes