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)))

Innermost 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

         ↳Nar

       →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)))

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 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →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)))

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 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

       →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)))

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 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

         ↳Nar

       →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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

       →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)))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, N, add(M, X)) -> 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

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

RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))

four new Dependency Pairs are created:

RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 8

                 ↳Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, N, add(M, X)) -> RM(N, X)
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, 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

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

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

three new Dependency Pairs are created:

IFRM(true, 0, add(0, X'')) -> RM(0, X'')
IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')
IFRM(true, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, 0, add(0, X'')) -> RM(0, X'')
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, N, add(M, X)) -> RM(N, X)
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), 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

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

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

five new Dependency Pairs are created:

IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
IFRM(true, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), 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

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

IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')

three new Dependency Pairs are created:

IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), 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

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

IFRM(true, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')

three new Dependency Pairs are created:

IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

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

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

IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')

five new Dependency Pairs are created:

IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

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

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

RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))

five new Dependency Pairs are created:

RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), X''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), X''''')))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), 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

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

IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')

five new Dependency Pairs are created:

IFRM(false, s(0), add(s(s(X''''')), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(0, X'0'''')))) -> RM(s(0), add(0, add(0, X'0'''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(0), X''''''')))) -> RM(s(0), add(0, add(s(0), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(s(X'''''''''')), X''''''')))) -> RM(s(0), add(0, add(s(s(X'''''''''')), X''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), X''''')))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(s(X'''''''''')), X''''''')))) -> RM(s(0), add(0, add(s(s(X'''''''''')), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(0), X''''''')))) -> RM(s(0), add(0, add(s(0), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(0, X'0'''')))) -> RM(s(0), add(0, add(0, X'0'''')))
IFRM(false, s(0), add(s(s(X''''')), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), 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

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

IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')

five new Dependency Pairs are created:

IFRM(false, s(s(X'''''')), add(s(0), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(false, s(s(X'''''')), add(s(0), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(0), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X'''''')), add(s(0), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), X''''')))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(s(X'''''''''')), X''''''')))) -> RM(s(0), add(0, add(s(s(X'''''''''')), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(0), X''''''')))) -> RM(s(0), add(0, add(s(0), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(0, X'0'''')))) -> RM(s(0), add(0, add(0, X'0'''')))
IFRM(false, s(0), add(s(s(X''''')), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X'''''')), add(s(0), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(0), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), 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

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

IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')

five new Dependency Pairs are created:

IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))
IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X'''''')), add(s(0), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X'''''')), add(s(0), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(0), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(s(X'''''''''')), X''''''')))) -> RM(s(0), add(0, add(s(s(X'''''''''')), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(0), X''''''')))) -> RM(s(0), add(0, add(s(0), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(0, X'0'''')))) -> RM(s(0), add(0, add(0, X'0'''')))
IFRM(false, s(0), add(s(s(X''''')), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), X''''')))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), 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

The following dependency pairs can be strictly oriented:

IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))
IFRM(false, s(s(X''''''')), add(s(s(Y'''')), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(0), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(0), X'''''''')))
IFRM(false, s(s(X'''''')), add(s(0), add(0, add(0, X'0'''')))) -> RM(s(s(X'''''')), add(0, add(0, X'0'''')))
IFRM(false, s(s(X'''''')), add(s(0), add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
IFRM(false, s(s(X'''''')), add(s(0), add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
IFRM(true, s(0), add(s(0), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(true, s(0), add(s(0), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(s(X'''''''''')), X''''''')))) -> RM(s(0), add(0, add(s(s(X'''''''''')), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(s(0), X''''''')))) -> RM(s(0), add(0, add(s(0), X''''''')))
IFRM(false, s(0), add(s(s(X''''')), add(0, add(0, X'0'''')))) -> RM(s(0), add(0, add(0, X'0'''')))
IFRM(false, s(0), add(s(s(X''''')), add(s(s(X''''')), X'''))) -> RM(s(0), add(s(s(X''''')), X'''))
IFRM(false, s(0), add(s(s(X''''')), add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
IFRM(true, s(0), add(s(0), add(0, X'0))) -> RM(s(0), add(0, X'0))
IFRM(true, s(s(X'''''')), add(s(s(Y'''')), add(0, X'0))) -> RM(s(s(X'''''')), add(0, X'0))
IFRM(false, s(s(X''''''')), add(s(0), add(0, add(s(s(Y'''''''')), X'''''''')))) -> RM(s(s(X''''''')), add(0, add(s(s(Y'''''''')), X'''''''')))

Additionally, the following usable rules for innermost can be oriented:

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

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)) = 1

POL(RM(x₁, x₂)) = x₂

POL(add(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 24

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(s(s(X''')), add(s(0), X)) -> IFRM(false, s(s(X''')), add(s(0), X))
IFRM(false, s(s(X'''''')), add(0, add(s(0), X'''))) -> RM(s(s(X'''''')), add(s(0), X'''))
RM(s(s(X'''''''')), add(0, add(s(0), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(0), X''''')))
RM(s(0), add(s(s(X''')), X)) -> IFRM(false, s(0), add(s(s(X''')), X))
IFRM(false, s(0), add(0, add(s(s(X'''''')), X'''))) -> RM(s(0), add(s(s(X'''''')), X'''))
RM(s(0), add(0, add(s(s(X'''''''')), X'''''))) -> IFRM(false, s(0), add(0, add(s(s(X'''''''')), X''''')))
RM(s(0), add(s(0), X)) -> IFRM(true, s(0), add(s(0), X))
IFRM(false, s(0), add(0, add(s(0), X'''))) -> RM(s(0), add(s(0), X'''))
RM(s(0), add(0, add(s(0), X'''''))) -> IFRM(false, s(0), add(0, add(s(0), X''''')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))
RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
RM(s(s(X''')), add(s(s(Y'')), X)) -> IFRM(eq(X''', Y''), s(s(X''')), add(s(s(Y'')), X))
IFRM(false, s(s(X'''''')), add(0, add(s(s(Y'''')), X'''))) -> RM(s(s(X'''''')), add(s(s(Y'''')), X'''))
RM(s(s(X'''''''')), add(0, add(s(s(Y'''''')), X'''''))) -> IFRM(false, s(s(X'''''''')), add(0, add(s(s(Y'''''')), 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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 26

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))
IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))

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

The following dependency pair can be strictly oriented:

RM(s(X'''), add(0, add(0, X'0''))) -> IFRM(false, s(X'''), add(0, add(0, X'0'')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IFRM(x₁, x₂, x₃)) = x₃

POL(0) = 0

POL(false) = 0

POL(s(x₁)) = 0

POL(RM(x₁, x₂)) = 1 + x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 28

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

IFRM(false, s(X'''''), add(0, add(0, X'0))) -> RM(s(X'''''), add(0, X'0))

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, 0, add(0, X'')) -> RM(0, X'')
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), 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

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

IFRM(true, 0, add(0, X'')) -> RM(0, X'')

two new Dependency Pairs are created:

IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
RM(0, add(0, X)) -> IFRM(true, 0, add(0, 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

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

RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))

two new Dependency Pairs are created:

RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
RM(0, add(0, add(s(X''''''), X'0''))) -> IFRM(true, 0, add(0, add(s(X''''''), X'0'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
RM(0, add(0, add(s(X''''''), X'0''))) -> IFRM(true, 0, add(0, add(s(X''''''), X'0'')))
IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(false, 0, add(s(X''''), X'')) -> RM(0, 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

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

IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')

three new Dependency Pairs are created:

IFRM(false, 0, add(s(X''''), add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
IFRM(false, 0, add(s(X''''), add(0, add(0, X''''''')))) -> RM(0, add(0, add(0, X''''''')))
IFRM(false, 0, add(s(X''''), add(0, add(s(X''''''''), X'0'''')))) -> RM(0, add(0, add(s(X''''''''), X'0'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(false, 0, add(s(X''''), add(0, add(s(X''''''''), X'0'''')))) -> RM(0, add(0, add(s(X''''''''), X'0'''')))
RM(0, add(0, add(s(X''''''), X'0''))) -> IFRM(true, 0, add(0, add(s(X''''''), X'0'')))
IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
IFRM(false, 0, add(s(X''''), add(0, add(0, X''''''')))) -> RM(0, add(0, add(0, X''''''')))
IFRM(false, 0, add(s(X''''), add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))

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

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

RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))

three new Dependency Pairs are created:

RM(0, add(s(X'''), add(s(X'''''''), X'0''))) -> IFRM(false, 0, add(s(X'''), add(s(X'''''''), X'0'')))
RM(0, add(s(X'''), add(0, add(0, X''''''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(0, X'''''''''))))
RM(0, add(s(X'''), add(0, add(s(X''''''''''), X'0'''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(s(X''''''''''), X'0''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 19

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(0, add(s(X'''), add(0, add(s(X''''''''''), X'0'''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(s(X''''''''''), X'0''''''))))
IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
IFRM(false, 0, add(s(X''''), add(0, add(0, X''''''')))) -> RM(0, add(0, add(0, X''''''')))
RM(0, add(s(X'''), add(0, add(0, X''''''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(0, X'''''''''))))
IFRM(false, 0, add(s(X''''), add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
RM(0, add(s(X'''), add(s(X'''''''), X'0''))) -> IFRM(false, 0, add(s(X'''), add(s(X'''''''), X'0'')))
IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
RM(0, add(0, add(s(X''''''), X'0''))) -> IFRM(true, 0, add(0, add(s(X''''''), X'0'')))
IFRM(false, 0, add(s(X''''), add(0, add(s(X''''''''), X'0'''')))) -> RM(0, add(0, add(s(X''''''''), X'0'''')))

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

The following dependency pairs can be strictly oriented:

RM(0, add(s(X'''), add(0, add(s(X''''''''''), X'0'''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(s(X''''''''''), X'0''''''))))
RM(0, add(s(X'''), add(0, add(0, X''''''''')))) -> IFRM(false, 0, add(s(X'''), add(0, add(0, X'''''''''))))
RM(0, add(s(X'''), add(s(X'''''''), X'0''))) -> IFRM(false, 0, add(s(X'''), add(s(X'''''''), X'0'')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IFRM(x₁, x₂, x₃)) = x₁ + x₃

POL(0) = 0

POL(false) = 0

POL(true) = 1

POL(s(x₁)) = 1

POL(RM(x₁, x₂)) = 1 + x₂

POL(add(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, X'''))
RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
IFRM(false, 0, add(s(X''''), add(0, add(0, X''''''')))) -> RM(0, add(0, add(0, X''''''')))
IFRM(false, 0, add(s(X''''), add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
IFRM(true, 0, add(0, add(s(X''''), X'0))) -> RM(0, add(s(X''''), X'0))
RM(0, add(0, add(s(X''''''), X'0''))) -> IFRM(true, 0, add(0, add(s(X''''''), X'0'')))
IFRM(false, 0, add(s(X''''), add(0, add(s(X''''''''), X'0'''')))) -> RM(0, add(0, add(s(X''''''''), X'0'''')))

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))
IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, 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

The following dependency pair can be strictly oriented:

RM(0, add(0, add(0, X'''''))) -> IFRM(true, 0, add(0, add(0, X''''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IFRM(x₁, x₂, x₃)) = x₃

POL(0) = 0

POL(true) = 0

POL(RM(x₁, x₂)) = 1 + x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 27

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

IFRM(true, 0, add(0, add(0, X'''))) -> RM(0, add(0, 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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →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)))

Strategy:

innermost

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

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Narrowing Transformation

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)))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

PURGE(add(0, add(0, X''))) -> PURGE(ifrm(true, 0, add(0, X'')))
PURGE(add(0, add(s(X'), X''))) -> PURGE(ifrm(false, 0, add(s(X'), X'')))
PURGE(add(s(X'), add(0, X''))) -> PURGE(ifrm(false, s(X'), add(0, X'')))
PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 30

                 ↳Rewriting Transformation

Dependency Pairs:

PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), X'')))
PURGE(add(s(X'), add(0, X''))) -> PURGE(ifrm(false, s(X'), add(0, X'')))
PURGE(add(0, add(s(X'), X''))) -> PURGE(ifrm(false, 0, add(s(X'), X'')))
PURGE(add(0, add(0, X''))) -> PURGE(ifrm(true, 0, add(0, 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

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

PURGE(add(0, add(0, X''))) -> PURGE(ifrm(true, 0, add(0, X'')))

one new Dependency Pair is created:

PURGE(add(0, add(0, X''))) -> PURGE(rm(0, X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 31

                 ↳Rewriting Transformation

Dependency Pairs:

PURGE(add(0, add(0, X''))) -> PURGE(rm(0, X''))
PURGE(add(s(X'), add(0, X''))) -> PURGE(ifrm(false, s(X'), add(0, X'')))
PURGE(add(0, add(s(X'), X''))) -> PURGE(ifrm(false, 0, add(s(X'), X'')))
PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), 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

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

PURGE(add(0, add(s(X'), X''))) -> PURGE(ifrm(false, 0, add(s(X'), X'')))

one new Dependency Pair is created:

PURGE(add(0, add(s(X'), X''))) -> PURGE(add(s(X'), rm(0, X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 32

                 ↳Rewriting Transformation

Dependency Pairs:

PURGE(add(0, add(s(X'), X''))) -> PURGE(add(s(X'), rm(0, X'')))
PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), X'')))
PURGE(add(s(X'), add(0, X''))) -> PURGE(ifrm(false, s(X'), add(0, X'')))
PURGE(add(0, add(0, X''))) -> PURGE(rm(0, 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

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

PURGE(add(s(X'), add(0, X''))) -> PURGE(ifrm(false, s(X'), add(0, X'')))

one new Dependency Pair is created:

PURGE(add(s(X'), add(0, X''))) -> PURGE(add(0, rm(s(X'), X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 33

                 ↳Polynomial Ordering

Dependency Pairs:

PURGE(add(s(X'), add(0, X''))) -> PURGE(add(0, rm(s(X'), X'')))
PURGE(add(0, add(0, X''))) -> PURGE(rm(0, X''))
PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), X'')))
PURGE(add(0, add(s(X'), X''))) -> PURGE(add(s(X'), rm(0, 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

The following dependency pairs can be strictly oriented:

PURGE(add(s(X'), add(0, X''))) -> PURGE(add(0, rm(s(X'), X'')))
PURGE(add(0, add(0, X''))) -> PURGE(rm(0, X''))
PURGE(add(s(X'), add(s(Y'), X''))) -> PURGE(ifrm(eq(X', Y'), s(X'), add(s(Y'), X'')))
PURGE(add(0, add(s(X'), X''))) -> PURGE(add(s(X'), rm(0, X'')))

Additionally, the following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(PURGE(x₁)) = 1 + x₁

POL(false) = 0

POL(nil) = 0

POL(ifrm(x₁, x₂, x₃)) = x₃

POL(true) = 0

POL(rm(x₁, x₂)) = x₂

POL(s(x₁)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 29

             ↳Nar

...

               →DP Problem 34

                 ↳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)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:06 minutes

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

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

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