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)))

Innermost 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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

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)))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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(EQ(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

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)))

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

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

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)))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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(LE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

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)))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

REPLACE(0, M, cons(0, L)) -> IFREPL(true, 0, M, cons(0, L))
REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), L))
REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, L))
REPLACE(s(X'), M, cons(s(Y'), L)) -> IFREPL(eq(X', Y'), s(X'), M, cons(s(Y'), L))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(X'), M, cons(s(Y'), L)) -> IFREPL(eq(X', Y'), s(X'), M, cons(s(Y'), L))
REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, L))
REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), L))
IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, 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)))

Strategy:

innermost

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

REPLACE(s(X'), M, cons(s(Y'), L)) -> IFREPL(eq(X', Y'), s(X'), M, cons(s(Y'), L))

four new Dependency Pairs are created:

REPLACE(s(0), M, cons(s(0), L)) -> IFREPL(true, s(0), M, cons(s(0), L))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 13

                 ↳Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), L))
IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, M, L)
REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, 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)))

Strategy:

innermost

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

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

five new Dependency Pairs are created:

IFREPL(false, 0, M'', cons(s(Y'''), L'')) -> REPLACE(0, M'', L'')
IFREPL(false, s(X'''), M'', cons(0, L'')) -> REPLACE(s(X'''), M'', L'')
IFREPL(false, s(0), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(0), M'', L'')
IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), M'', L'')
IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(0), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(0), M'', L'')
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(X'''), M'', cons(0, L'')) -> REPLACE(s(X'''), M'', L'')
REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, L))
IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), 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)))

Strategy:

innermost

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

IFREPL(false, s(X'''), M'', cons(0, L'')) -> REPLACE(s(X'''), M'', L'')

four new Dependency Pairs are created:

IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(0), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(0), M'', L'')
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, L))
IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), 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)))

Strategy:

innermost

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

REPLACE(s(X'), M, cons(0, L)) -> IFREPL(false, s(X'), M, cons(0, L))

four new Dependency Pairs are created:

REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
IFREPL(false, s(0), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(0), M'', L'')
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), 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)))

Strategy:

innermost

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

IFREPL(false, s(0), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(0), M'', L'')

three new Dependency Pairs are created:

IFREPL(false, s(0), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(0), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), 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)))

Strategy:

innermost

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

IFREPL(false, s(s(X'''')), M'', cons(s(0), L'')) -> REPLACE(s(s(X'''')), M'', L'')

five new Dependency Pairs are created:

IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(0), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), 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)))

Strategy:

innermost

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

IFREPL(false, s(s(X'''')), M'', cons(s(s(Y'''')), L'')) -> REPLACE(s(s(X'''')), M'', L'')

five new Dependency Pairs are created:

IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(0), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), 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)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(s(X''''')), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(0), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(0), L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))
IFREPL(false, s(0), M'''', cons(s(s(Y'''')), cons(0, cons(0, L''''''')))) -> REPLACE(s(0), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(0), M''', cons(s(s(Y'''')), cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(0, L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(0, L''''''')))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(0), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(s(X''''')), M''', cons(s(s(Y'''')), cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
IFREPL(false, s(s(X''''')), M'''', cons(s(0), cons(0, cons(s(s(Y'''''''')), L''''''')))) -> REPLACE(s(s(X''''')), M'''', cons(0, cons(s(s(Y'''''''')), L''''''')))

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(REPLACE(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(true) = 0

POL(s(x₁)) = 1

POL(IFREPL(x₁, x₂, x₃, x₄)) = x₄

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 24

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(0), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(0), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(0), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(0), L''''')))
REPLACE(s(0), M, cons(s(s(Y'')), L)) -> IFREPL(false, s(0), M, cons(s(s(Y'')), L))
IFREPL(false, s(0), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(0), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(0), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(0), M', cons(0, cons(s(s(Y'''''')), L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, L'''))
REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
REPLACE(s(s(X'')), M, cons(s(0), L)) -> IFREPL(false, s(s(X'')), M, cons(s(0), L))
REPLACE(s(s(X'')), M, cons(s(s(Y'')), L)) -> IFREPL(eq(X'', Y''), s(s(X'')), M, cons(s(s(Y'')), L))
IFREPL(false, s(s(X''''')), M''', cons(0, cons(s(s(Y'''')), L'''))) -> REPLACE(s(s(X''''')), M''', cons(s(s(Y'''')), L'''))
REPLACE(s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), L'''''))) -> IFREPL(false, s(s(X''''''')), M', cons(0, cons(s(s(Y'''''')), 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)))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))
IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, 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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REPLACE(s(X''), M', cons(0, cons(0, L'''''))) -> IFREPL(false, s(X''), M', cons(0, cons(0, L''''')))

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(REPLACE(x₁, x₂, x₃)) = 1 + x₃

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(false) = 0

POL(s(x₁)) = 0

POL(IFREPL(x₁, x₂, x₃, x₄)) = x₄

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 26

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

IFREPL(false, s(X''''), M''', cons(0, cons(0, L'''))) -> REPLACE(s(X''''), M''', cons(0, 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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, 0, M'', cons(s(Y'''), L'')) -> REPLACE(0, M'', L'')
REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), 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)))

Strategy:

innermost

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

IFREPL(false, 0, M'', cons(s(Y'''), L'')) -> REPLACE(0, M'', L'')

one new Dependency Pair is created:

IFREPL(false, 0, M''', cons(s(Y'''), cons(s(Y'''), L'''))) -> REPLACE(0, M''', cons(s(Y'''), L'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFREPL(false, 0, M''', cons(s(Y'''), cons(s(Y'''), L'''))) -> REPLACE(0, M''', cons(s(Y'''), L'''))
REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), 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)))

Strategy:

innermost

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

REPLACE(0, M, cons(s(Y'), L)) -> IFREPL(false, 0, M, cons(s(Y'), L))

one new Dependency Pair is created:

REPLACE(0, M', cons(s(Y''), cons(s(Y''''''), L'''''))) -> IFREPL(false, 0, M', cons(s(Y''), cons(s(Y''''''), L''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 19

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(0, M', cons(s(Y''), cons(s(Y''''''), L'''''))) -> IFREPL(false, 0, M', cons(s(Y''), cons(s(Y''''''), L''''')))
IFREPL(false, 0, M''', cons(s(Y'''), cons(s(Y'''), L'''))) -> REPLACE(0, M''', cons(s(Y'''), 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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REPLACE(0, M', cons(s(Y''), cons(s(Y''''''), L'''''))) -> IFREPL(false, 0, M', cons(s(Y''), cons(s(Y''''''), L''''')))

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(REPLACE(x₁, x₂, x₃)) = 1 + x₃

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(false) = 0

POL(s(x₁)) = 1

POL(IFREPL(x₁, x₂, x₃, x₄)) = x₄

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

IFREPL(false, 0, M''', cons(s(Y'''), cons(s(Y'''), L'''))) -> REPLACE(0, M''', cons(s(Y'''), 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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Narrowing Transformation

       →DP Problem 5

         ↳Nar

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)))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, cons(M', L)))
MIN(cons(s(X'), cons(0, L))) -> IFMIN(false, cons(s(X'), cons(0, L)))
MIN(cons(s(X'), cons(s(Y'), L))) -> IFMIN(le(X', Y'), cons(s(X'), cons(s(Y'), L)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Narrowing Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(s(X'), cons(s(Y'), L))) -> IFMIN(le(X', Y'), cons(s(X'), cons(s(Y'), L)))
MIN(cons(s(X'), cons(0, L))) -> IFMIN(false, cons(s(X'), cons(0, L)))
IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))
MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, cons(M', L)))
IFMIN(false, cons(N, cons(M, L))) -> MIN(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)))

Strategy:

innermost

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

MIN(cons(s(X'), cons(s(Y'), L))) -> IFMIN(le(X', Y'), cons(s(X'), cons(s(Y'), L)))

three new Dependency Pairs are created:

MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), L)))
MIN(cons(s(s(X'')), cons(s(0), L))) -> IFMIN(false, cons(s(s(X'')), cons(s(0), L)))
MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), L)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 28

                 ↳Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), L)))
MIN(cons(s(s(X'')), cons(s(0), L))) -> IFMIN(false, cons(s(s(X'')), cons(s(0), L)))
MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), L)))
IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))
MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, cons(M', L)))
IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
MIN(cons(s(X'), cons(0, L))) -> IFMIN(false, cons(s(X'), cons(0, 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)))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

IFMIN(true, cons(0, cons(M', L''))) -> MIN(cons(0, L''))
IFMIN(true, cons(s(0), cons(s(Y''''), L''))) -> MIN(cons(s(0), L''))
IFMIN(true, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(X'''')), L''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 29

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(0, cons(M', L''))) -> MIN(cons(0, L''))
MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, 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)))

Strategy:

innermost

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

IFMIN(true, cons(0, cons(M', L''))) -> MIN(cons(0, L''))

one new Dependency Pair is created:

IFMIN(true, cons(0, cons(M', cons(M''', L''')))) -> MIN(cons(0, cons(M''', L''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 31

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(0, cons(M', cons(M''', L''')))) -> MIN(cons(0, cons(M''', L''')))
MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, 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)))

Strategy:

innermost

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

MIN(cons(0, cons(M', L))) -> IFMIN(true, cons(0, cons(M', L)))

one new Dependency Pair is created:

MIN(cons(0, cons(M''', cons(M''''', L''''')))) -> IFMIN(true, cons(0, cons(M''', cons(M''''', L'''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 34

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(0, cons(M''', cons(M''''', L''''')))) -> IFMIN(true, cons(0, cons(M''', cons(M''''', L'''''))))
IFMIN(true, cons(0, cons(M', cons(M''', L''')))) -> MIN(cons(0, 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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MIN(cons(0, cons(M''', cons(M''''', L''''')))) -> IFMIN(true, cons(0, cons(M''', cons(M''''', L'''''))))

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(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(IFMIN(x₁, x₂)) = x₂

POL(MIN(x₁)) = 1 + x₁

POL(true) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 38

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

IFMIN(true, cons(0, cons(M', cons(M''', L''')))) -> MIN(cons(0, 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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 30

                 ↳Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(X'''')), L''))
MIN(cons(s(s(X'')), cons(s(0), L))) -> IFMIN(false, cons(s(s(X'')), cons(s(0), L)))
IFMIN(true, cons(s(0), cons(s(Y''''), L''))) -> MIN(cons(s(0), L''))
MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), L)))
MIN(cons(s(X'), cons(0, L))) -> IFMIN(false, cons(s(X'), cons(0, L)))
IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), 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)))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

IFMIN(false, cons(s(X'''), cons(0, L''))) -> MIN(cons(0, L''))
IFMIN(false, cons(s(s(X'''')), cons(s(0), L''))) -> MIN(cons(s(0), L''))
IFMIN(false, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(Y'''')), L''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 32

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(s(0), cons(s(Y''''), L''))) -> MIN(cons(s(0), L''))
MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), 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)))

Strategy:

innermost

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

IFMIN(true, cons(s(0), cons(s(Y''''), L''))) -> MIN(cons(s(0), L''))

one new Dependency Pair is created:

IFMIN(true, cons(s(0), cons(s(Y''''), cons(s(Y''''), L''')))) -> MIN(cons(s(0), cons(s(Y''''), L''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 35

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(s(0), cons(s(Y''''), cons(s(Y''''), L''')))) -> MIN(cons(s(0), cons(s(Y''''), L''')))
MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), 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)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMIN(true, cons(s(0), cons(s(Y''''), cons(s(Y''''), L''')))) -> MIN(cons(s(0), cons(s(Y''''), L''')))

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(0) = 0

POL(IFMIN(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = x₁ + x₂

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 39

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

MIN(cons(s(0), cons(s(Y''), L))) -> IFMIN(true, cons(s(0), cons(s(Y''), 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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 33

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(false, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(Y'''')), L''))
MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), L)))
IFMIN(true, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(X'''')), 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)))

Strategy:

innermost

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

IFMIN(true, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(X'''')), L''))

one new Dependency Pair is created:

IFMIN(true, cons(s(s(X''''')), cons(s(s(Y'''')), cons(s(s(Y'''')), L''')))) -> MIN(cons(s(s(X''''')), cons(s(s(Y'''')), L''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 36

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(true, cons(s(s(X''''')), cons(s(s(Y'''')), cons(s(s(Y'''')), L''')))) -> MIN(cons(s(s(X''''')), cons(s(s(Y'''')), L''')))
MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), L)))
IFMIN(false, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(Y'''')), 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)))

Strategy:

innermost

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

IFMIN(false, cons(s(s(X'''')), cons(s(s(Y'''')), L''))) -> MIN(cons(s(s(Y'''')), L''))

one new Dependency Pair is created:

IFMIN(false, cons(s(s(X'''')), cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))) -> MIN(cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 37

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IFMIN(false, cons(s(s(X'''')), cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))) -> MIN(cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))
MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), L)))
IFMIN(true, cons(s(s(X''''')), cons(s(s(Y'''')), cons(s(s(Y'''')), L''')))) -> MIN(cons(s(s(X''''')), cons(s(s(Y'''')), 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)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IFMIN(false, cons(s(s(X'''')), cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))) -> MIN(cons(s(s(Y''''')), cons(s(s(Y'''0)), L''')))
IFMIN(true, cons(s(s(X''''')), cons(s(s(Y'''')), cons(s(s(Y'''')), L''')))) -> MIN(cons(s(s(X''''')), cons(s(s(Y'''')), L''')))

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(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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 27

             ↳Nar

...

               →DP Problem 40

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

MIN(cons(s(s(X'')), cons(s(s(Y'')), L))) -> IFMIN(le(X'', Y''), cons(s(s(X'')), cons(s(s(Y'')), 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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Narrowing Transformation

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)))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

SELSORT(cons(0, nil)) -> IFSELSORT(eq(0, 0), cons(0, nil))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(s(N''), s(N'')), cons(s(N''), nil))
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rewriting Transformation

Dependency Pairs:

SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(s(N''), s(N'')), cons(s(N''), nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(0, nil)) -> IFSELSORT(eq(0, 0), cons(0, nil))
IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), 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)))

Strategy:

innermost

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

SELSORT(cons(0, nil)) -> IFSELSORT(eq(0, 0), cons(0, nil))

one new Dependency Pair is created:

SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 42

                 ↳Rewriting Transformation

Dependency Pairs:

SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), N, L))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(s(N''), s(N'')), cons(s(N''), nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), 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)))

Strategy:

innermost

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

SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(s(N''), s(N'')), cons(s(N''), nil))

one new Dependency Pair is created:

SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 43

                 ↳Narrowing Transformation

Dependency Pairs:

SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
IFSELSORT(false, cons(N, L)) -> SELSORT(replace(min(cons(N, L)), N, L))
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))

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)))

Strategy:

innermost

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

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

five new Dependency Pairs are created:

IFSELSORT(false, cons(N'', nil)) -> SELSORT(nil)
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(min(cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))
IFSELSORT(false, cons(0, nil)) -> SELSORT(replace(0, 0, nil))
IFSELSORT(false, cons(s(N''), nil)) -> SELSORT(replace(s(N''), s(N''), nil))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(replace(ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 44

                 ↳Rewriting Transformation

Dependency Pairs:

IFSELSORT(false, cons(s(N''), nil)) -> SELSORT(replace(s(N''), s(N''), nil))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(replace(ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(min(cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))

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)))

Strategy:

innermost

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

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(min(cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))

one new Dependency Pair is created:

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 45

                 ↳Rewriting Transformation

Dependency Pairs:

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(replace(ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(false, cons(s(N''), nil)) -> SELSORT(replace(s(N''), s(N''), nil))

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)))

Strategy:

innermost

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

IFSELSORT(false, cons(s(N''), nil)) -> SELSORT(replace(s(N''), s(N''), nil))

one new Dependency Pair is created:

IFSELSORT(false, cons(s(N''), nil)) -> SELSORT(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 46

                 ↳Rewriting Transformation

Dependency Pairs:

IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(replace(ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', 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)))

Strategy:

innermost

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

IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(replace(ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))

one new Dependency Pair is created:

IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 47

                 ↳Rewriting Transformation

Dependency Pairs:

IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))

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)))

Strategy:

innermost

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

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), min(cons(N'', cons(K', L''))), N'', cons(K', L'')))

one new Dependency Pair is created:

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 48

                 ↳Narrowing Transformation

Dependency Pairs:

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), 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)))

Strategy:

innermost

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

SELSORT(cons(N'', cons(M', L''))) -> IFSELSORT(eq(N'', ifmin(le(N'', M'), cons(N'', cons(M', L'')))), cons(N'', cons(M', L'')))

three new Dependency Pairs are created:

SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, ifmin(true, cons(0, cons(M'', L'')))), cons(0, cons(M'', L'')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), ifmin(false, cons(s(X'), cons(0, L'')))), cons(s(X'), cons(0, L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 49

                 ↳Rewriting Transformation

Dependency Pairs:

SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), ifmin(false, cons(s(X'), cons(0, L'')))), cons(s(X'), cons(0, L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, ifmin(true, cons(0, cons(M'', L'')))), cons(0, cons(M'', L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', 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)))

Strategy:

innermost

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

SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, ifmin(true, cons(0, cons(M'', L'')))), cons(0, cons(M'', L'')))

one new Dependency Pair is created:

SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 50

                 ↳Rewriting Transformation

Dependency Pairs:

IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), ifmin(false, cons(s(X'), cons(0, L'')))), cons(s(X'), cons(0, L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), 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)))

Strategy:

innermost

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

SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), ifmin(false, cons(s(X'), cons(0, L'')))), cons(s(X'), cons(0, L'')))

one new Dependency Pair is created:

SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, L'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 51

                 ↳Instantiation Transformation

Dependency Pairs:

SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, L'')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
SELSORT(cons(s(N''), nil)) -> IFSELSORT(eq(N'', N''), cons(s(N''), nil))
IFSELSORT(true, cons(N, L)) -> SELSORT(L)
SELSORT(cons(0, nil)) -> IFSELSORT(true, cons(0, nil))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', 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)))

Strategy:

innermost

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

IFSELSORT(true, cons(N, L)) -> SELSORT(L)

five new Dependency Pairs are created:

IFSELSORT(true, cons(0, nil)) -> SELSORT(nil)
IFSELSORT(true, cons(s(N''''), nil)) -> SELSORT(nil)
IFSELSORT(true, cons(s(X'''), cons(s(Y'''), L''''))) -> SELSORT(cons(s(Y'''), L''''))
IFSELSORT(true, cons(0, cons(M'''', L''''))) -> SELSORT(cons(M'''', L''''))
IFSELSORT(true, cons(s(X'''), cons(0, L''''))) -> SELSORT(cons(0, L''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 52

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IFSELSORT(true, cons(s(X'''), cons(0, L''''))) -> SELSORT(cons(0, L''''))
IFSELSORT(true, cons(s(X'''), cons(s(Y'''), L''''))) -> SELSORT(cons(s(Y'''), L''''))
IFSELSORT(true, cons(0, cons(M'''', L''''))) -> SELSORT(cons(M'''', L''''))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, 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)))

Strategy:

innermost

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

IFSELSORT(true, cons(s(X'''), cons(s(Y'''), L''''))) -> SELSORT(cons(s(Y'''), L''''))

two new Dependency Pairs are created:

IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(s(Y''0), L''''')))) -> SELSORT(cons(s(Y''''), cons(s(Y''0), L''''')))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(0, L''''')))) -> SELSORT(cons(s(Y''''), cons(0, L''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 53

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IFSELSORT(true, cons(0, cons(M'''', L''''))) -> SELSORT(cons(M'''', L''''))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(0, L''''')))) -> SELSORT(cons(s(Y''''), cons(0, L''''')))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(s(Y''0), L''''')))) -> SELSORT(cons(s(Y''''), cons(s(Y''0), L''''')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, L'')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(true, cons(s(X'''), cons(0, L''''))) -> SELSORT(cons(0, 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)))

Strategy:

innermost

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

IFSELSORT(true, cons(0, cons(M'''', L''''))) -> SELSORT(cons(M'''', L''''))

three new Dependency Pairs are created:

IFSELSORT(true, cons(0, cons(s(X'''), cons(s(Y'''), L''''')))) -> SELSORT(cons(s(X'''), cons(s(Y'''), L''''')))
IFSELSORT(true, cons(0, cons(0, cons(M''''', L''''')))) -> SELSORT(cons(0, cons(M''''', L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(0, L''''')))) -> SELSORT(cons(s(X'''), cons(0, L''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 54

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IFSELSORT(true, cons(s(X'''), cons(0, L''''))) -> SELSORT(cons(0, L''''))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(s(Y''0), L''''')))) -> SELSORT(cons(s(Y''''), cons(s(Y''0), L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(0, L''''')))) -> SELSORT(cons(s(X'''), cons(0, L''''')))
IFSELSORT(true, cons(0, cons(0, cons(M''''', L''''')))) -> SELSORT(cons(0, cons(M''''', L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(s(Y'''), L''''')))) -> SELSORT(cons(s(X'''), cons(s(Y'''), L''''')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, L'')))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(0, L''''')))) -> SELSORT(cons(s(Y''''), cons(0, 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)))

Strategy:

innermost

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

IFSELSORT(true, cons(s(X'''), cons(0, L''''))) -> SELSORT(cons(0, L''''))

one new Dependency Pair is created:

IFSELSORT(true, cons(s(X'''), cons(0, cons(M'''', L''''')))) -> SELSORT(cons(0, cons(M'''', L''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 55

                 ↳Polynomial Ordering

Dependency Pairs:

IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(0, L''''')))) -> SELSORT(cons(s(Y''''), cons(0, L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(0, L''''')))) -> SELSORT(cons(s(X'''), cons(0, L''''')))
IFSELSORT(true, cons(0, cons(0, cons(M''''', L''''')))) -> SELSORT(cons(0, cons(M''''', L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(s(Y'''), L''''')))) -> SELSORT(cons(s(X'''), cons(s(Y'''), L''''')))
IFSELSORT(true, cons(s(X'''), cons(0, cons(M'''', L''''')))) -> SELSORT(cons(0, cons(M'''', L''''')))
SELSORT(cons(s(X'), cons(0, L''))) -> IFSELSORT(eq(s(X'), min(cons(0, L''))), cons(s(X'), cons(0, L'')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
SELSORT(cons(0, cons(M'', L''))) -> IFSELSORT(eq(0, min(cons(0, L''))), cons(0, cons(M'', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
SELSORT(cons(s(X'), cons(s(Y'), L''))) -> IFSELSORT(eq(s(X'), ifmin(le(X', Y'), cons(s(X'), cons(s(Y'), L'')))), cons(s(X'), cons(s(Y'), L'')))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(s(Y''0), L''''')))) -> SELSORT(cons(s(Y''''), cons(s(Y''0), 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)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(0, L''''')))) -> SELSORT(cons(s(Y''''), cons(0, L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(0, L''''')))) -> SELSORT(cons(s(X'''), cons(0, L''''')))
IFSELSORT(true, cons(0, cons(0, cons(M''''', L''''')))) -> SELSORT(cons(0, cons(M''''', L''''')))
IFSELSORT(true, cons(0, cons(s(X'''), cons(s(Y'''), L''''')))) -> SELSORT(cons(s(X'''), cons(s(Y'''), L''''')))
IFSELSORT(true, cons(s(X'''), cons(0, cons(M'''', L''''')))) -> SELSORT(cons(0, cons(M'''', L''''')))
IFSELSORT(false, cons(N'', cons(K', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', K'), cons(N'', cons(K', L''))), K'), ifmin(le(N'', K'), cons(N'', cons(K', L''))), N'', cons(K', L'')))
IFSELSORT(false, cons(N'', cons(M', L''))) -> SELSORT(ifrepl(eq(ifmin(le(N'', M'), cons(N'', cons(M', L''))), M'), ifmin(le(N'', M'), cons(N'', cons(M', L''))), N'', cons(M', L'')))
IFSELSORT(true, cons(s(X'''), cons(s(Y''''), cons(s(Y''0), L''''')))) -> SELSORT(cons(s(Y''''), cons(s(Y''0), L''''')))

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

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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(false) = 0

POL(true) = 0

POL(replace(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 41

             ↳Rw

...

               →DP Problem 56

                 ↳Dependency Graph

Dependency Pairs:

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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:11 minutes

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

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

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