Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Overlay and local confluence Check

The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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:

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

D_P: 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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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:

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

D_P: 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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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:

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))

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

D_P: 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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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:

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)))

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

D_P: 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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

           →DP Problem 5

             ↳UsableRules

...

               →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.

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