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. 



   R
     ↳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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


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




As we are in the innermost case, we can delete all 20 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 6
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. EQ(s(X), s(Y)) -> EQ(X, Y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


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




As we are in the innermost case, we can delete all 20 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 7
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LE(s(X), s(Y)) -> LE(X, Y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


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




As we are in the innermost case, we can delete all 16 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 8
             ↳Size-Change Principle
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules


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), s(Y)) -> eq(X, Y)
eq(s(X), 0) -> false


Strategy:

innermost




We number the DPs as follows:
  1. IFREPL(false, N, M, cons(K, L)) -> REPLACE(N, M, L)
  2. REPLACE(N, M, cons(K, L)) -> IFREPL(eq(N, K), N, M, cons(K, L))
and get the following Size-Change Graph(s):
{1} , {1}
2=1
3=2
4>3
{2} , {2}
1=2
2=3
3=4

which lead(s) to this/these maximal multigraph(s):
{1} , {2}
2=2
3=3
4>4
{2} , {1}
1=1
2=2
3>3

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳Usable Rules (Innermost)
       →DP Problem 5
         ↳UsableRules


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




As we are in the innermost case, we can delete all 17 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 9
             ↳Size-Change Principle
       →DP Problem 5
         ↳UsableRules


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:


le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)


Strategy:

innermost




We number the DPs as follows:
  1. IFMIN(false, cons(N, cons(M, L))) -> MIN(cons(M, L))
  2. IFMIN(true, cons(N, cons(M, L))) -> MIN(cons(N, L))
  3. MIN(cons(N, cons(M, L))) -> IFMIN(le(N, M), cons(N, cons(M, L)))
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
2>1
{3} , {3}
1=2

which lead(s) to this/these maximal multigraph(s):
{3} , {1, 2}
1>1
{1, 2} , {3}
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
           →DP Problem 10
             ↳Negative Polynomial Order


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:


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


Strategy:

innermost




The following Dependency Pairs can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( IFSELSORT(x1, x2) ) = x2

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

POL( SELSORT(x1) ) = x1

POL( replace(x1, ..., x3) ) = x3

POL( nil ) = 0

POL( ifrepl(x1, ..., x4) ) = x4

POL( ifmin(x1, x2) ) = 0

POL( min(x1) ) = 0

POL( s(x1) ) = 0

POL( 0 ) = 0

POL( le(x1, x2) ) = 0

POL( false ) = 0

POL( true ) = 0

POL( eq(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
           →DP Problem 10
             ↳Neg POLO
             ...
               →DP Problem 11
                 ↳Dependency Graph


Dependency Pair:

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


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes