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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

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)))

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 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

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 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳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(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)))

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 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 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

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)))

The following dependency pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IFRM(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(true) = 0

POL(s(x₁)) = 0

POL(RM(x₁, x₂)) = x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

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)))

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

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

two new Dependency Pairs are created:

PURGE(add(N'', nil)) -> PURGE(nil)
PURGE(add(N'', add(M', X''))) -> PURGE(ifrm(eq(N'', M'), N'', add(M', X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 8

             ↳Polynomial Ordering

Dependency Pair:

PURGE(add(N'', add(M', X''))) -> PURGE(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)))

The following dependency pair can be strictly oriented:

PURGE(add(N'', add(M', X''))) -> PURGE(ifrm(eq(N'', M'), N'', add(M', X'')))

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

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 ordering with Polynomial interpretation:

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(PURGE(x₁)) = x₁

POL(false) = 0

POL(true) = 0

POL(ifrm(x₁, x₂, x₃)) = x₃

POL(nil) = 0

POL(s(x₁)) = 0

POL(rm(x₁, x₂)) = x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

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

POL(IFRM(x₁, x₂, x₃))	= x₃
POL(eq(x₁, x₂))	= 0
POL(0)	= 0
POL(false)	= 0
POL(true)	= 0
POL(s(x₁))	= 0
POL(RM(x₁, x₂))	= x₂
POL(add(x₁, x₂))	= 1 + x₂

POL(eq(x₁, x₂))	= 0
POL(0)	= 0
POL(PURGE(x₁))	= x₁
POL(false)	= 0
POL(true)	= 0
POL(ifrm(x₁, x₂, x₃))	= x₃
POL(nil)	= 0
POL(s(x₁))	= 0
POL(rm(x₁, x₂))	= x₂
POL(add(x₁, x₂))	= 1 + x₂