Termination Proof using AProVE

Term Rewriting System R:
[y, x, z]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
EQ(s(x), s(y)) -> EQ(x, y)
MINSORT(cons(x, y)) -> MIN(x, y)
MINSORT(cons(x, y)) -> MINSORT(del(min(x, y), cons(x, y)))
MINSORT(cons(x, y)) -> DEL(min(x, y), cons(x, y))
MIN(x, cons(y, z)) -> IF(le(x, y), min(x, z), min(y, z))
MIN(x, cons(y, z)) -> LE(x, y)
MIN(x, cons(y, z)) -> MIN(x, z)
MIN(x, cons(y, z)) -> MIN(y, z)
DEL(x, cons(y, z)) -> IF(eq(x, y), z, cons(y, del(x, z)))
DEL(x, cons(y, z)) -> EQ(x, y)
DEL(x, cons(y, z)) -> DEL(x, z)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

EQ(s(x), s(y)) -> EQ(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

The following dependency pair can be strictly oriented:

EQ(s(x), s(y)) -> EQ(x, y)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(EQ(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

The following dependency pairs can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = 1 + x₂

POL(MIN(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(DEL(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

MINSORT(cons(x, y)) -> MINSORT(del(min(x, y), cons(x, y)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Termination of R could not be shown.
Duration:
0:00 minutes

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

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