Termination Proof using AProVE

Term Rewriting System R:
[Y, X, N, M, L, K]
eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
LE(s(X), s(Y)) -> LE(X, Y)
MIN(cons(N, cons(M, L))) -> IFMIN(le(N, M), cons(N, cons(M, L)))
MIN(cons(N, cons(M, L))) -> LE(N, M)
IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
REPLACE(N, M, cons(K, L)) -> IFREPL(eq(N, K), N, M, cons(K, L))
REPLACE(N, M, cons(K, L)) -> EQ(N, K)
IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, M, L)
SELSORT(cons(N, L)) -> IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
SELSORT(cons(N, L)) -> EQ(N, min(cons(N, L)))
SELSORT(cons(N, L)) -> MIN(cons(N, L))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
IFSELSORT(false, cons(N, L)) -> MIN(cons(N, L))
IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(false, cons(N, L)) -> REPLACE(min(cons(N, L)), N, L)

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

         ↳Polo

Dependency Pair:

EQ(s(X), s(Y)) -> EQ(X, Y)

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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 6

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

         ↳Polo

Dependency Pair:

LE(s(X), s(Y)) -> LE(X, Y)

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

         ↳Polo

Dependency Pairs:

IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, M, L)
REPLACE(N, M, cons(K, L)) -> IFREPL(eq(N, K), N, M, cons(K, L))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

The following dependency pair can be strictly oriented:

IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, M, L)

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(REPLACE(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(true) = 0

POL(s(x₁)) = 0

POL(IFREPL(x₁, x₂, x₃, x₄)) = x₄

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

         ↳Polo

Dependency Pair:

REPLACE(N, M, cons(K, L)) -> IFREPL(eq(N, K), N, M, cons(K, L))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

         ↳Polo

Dependency Pairs:

IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))
MIN(cons(N, cons(M, L))) -> IFMIN(le(N, M), cons(N, cons(M, L)))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

The following dependency pairs can be strictly oriented:

IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))

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(0) = 0

POL(false) = 0

POL(IFMIN(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = 1 + x₂

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

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

         ↳Polo

Dependency Pair:

MIN(cons(N, cons(M, L))) -> IFMIN(le(N, M), cons(N, cons(M, L)))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

         ↳Polynomial Ordering

Dependency Pairs:

IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(N, L)) -> IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

The following dependency pairs can be strictly oriented:

IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(false) = 0

POL(true) = 0

POL(replace(x₁, x₂, x₃)) = x₃

POL(0) = 0

POL(eq(x₁, x₂)) = 0

POL(SELSORT(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₂

POL(IFSELSORT(x₁, x₂)) = x₂

POL(min(x₁)) = 0

POL(nil) = 0

POL(s(x₁)) = 0

POL(ifrepl(x₁, x₂, x₃, x₄)) = x₄

POL(le(x₁, x₂)) = 0

POL(ifmin(x₁, x₂)) = 0

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 5

         ↳Polo

           →DP Problem 10

             ↳Dependency Graph

Dependency Pair:

SELSORT(cons(N, L)) -> IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))

Rules:

eq(0, 0) -> true
eq(0, s(Y)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
min(cons(0, nil)) -> 0
min(cons(s(N), nil)) -> s(N)
min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
replace(N, M, nil) -> nil
replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
selsort(nil) -> nil
selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Using the Dependency Graph resulted in no new DP problems.

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

POL(REPLACE(x₁, x₂, x₃))	= x₃
POL(eq(x₁, x₂))	= 0
POL(0)	= 0
POL(false)	= 0
POL(cons(x₁, x₂))	= 1 + x₂
POL(true)	= 0
POL(s(x₁))	= 0
POL(IFREPL(x₁, x₂, x₃, x₄))	= x₄

POL(0)	= 0
POL(false)	= 0
POL(IFMIN(x₁, x₂))	= x₂
POL(cons(x₁, x₂))	= 1 + x₂
POL(MIN(x₁))	= x₁
POL(true)	= 0
POL(s(x₁))	= 0
POL(le(x₁, x₂))	= 0