Termination Proof using AProVE

Term Rewriting System R:
[X, Y, N, M]
eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

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)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))
RM(N, add(M, X)) -> EQ(N, M)
IFRM(true, N, add(M, X)) -> RM(N, X)
IFRM(false, N, add(M, X)) -> RM(N, X)
PURGE(add(N, X)) -> PURGE(rm(N, X))
PURGE(add(N, X)) -> RM(N, X)

Furthermore, R contains three SCCs.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Usable Rules (Innermost)

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

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

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

...

               →DP Problem 4

                 ↳Size-Change Principle

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳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

Dependency Pairs:

IFRM(false, N, add(M, X)) -> RM(N, X)
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

As we are in the innermost case, we can delete all 6 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

...

               →DP Problem 5

                 ↳Size-Change Principle

           →DP Problem 3

             ↳UsableRules

Dependency Pairs:

IFRM(false, N, add(M, X)) -> RM(N, X)
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))

Rules:

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

Strategy:

innermost

We number the DPs as follows:

IFRM(false, N, add(M, X)) -> RM(N, X)
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))

and get the following Size-Change Graph(s):

{1, 2}	,	{1, 2}
2	=	1
3	>	2

{3}	,	{3}
1	=	2
2	=	3

which lead(s) to this/these maximal multigraph(s):

{3}	,	{1, 2}
1	=	1
2	>	2

{1, 2}	,	{3}
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:

add(x₁, x₂) -> add(x₁, x₂)

We obtain no new DP problems.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳Usable Rules (Innermost)

Dependency Pair:

PURGE(add(N, X)) -> PURGE(rm(N, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 6

                 ↳Negative Polynomial Order

Dependency Pair:

PURGE(add(N, X)) -> PURGE(rm(N, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), s(Y)) -> eq(X, Y)
eq(s(X), 0) -> false
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))

Strategy:

innermost

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

PURGE(add(N, X)) -> PURGE(rm(N, X))

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

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), s(Y)) -> eq(X, Y)
eq(s(X), 0) -> false
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))

Used ordering:
Polynomial Order with Interpretation:

POL( PURGE(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₂ + 1

POL( rm(x₁, x₂) ) = x₂

POL( eq(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( ifrm(x₁, ..., x₃) ) = x₃

POL( nil ) = 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 7

                 ↳Dependency Graph

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), s(Y)) -> eq(X, Y)
eq(s(X), 0) -> false
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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