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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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)
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)

Furthermore, R contains three SCCs.

`   R`
`     ↳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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

`   R`
`     ↳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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

`   R`
`     ↳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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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 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.

`   R`
`     ↳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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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)

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

four new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pairs:

IFRM(true, N, add(M, X)) -> RM(N, 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

four new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pairs:

IFRM(true, N, add(M, X)) -> RM(N, 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

`   R`
`     ↳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'')
IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')
IFRM(true, 0, add(0, X'')) -> RM(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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

`   R`
`     ↳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'')
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
IFRM(true, s(0), add(s(0), X'')) -> RM(s(0), X'')
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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'')
IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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:

IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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:

IFRM(false, s(s(X''''')), add(s(0), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(0), add(s(s(X''''')), X'')) -> RM(s(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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

five new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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(false, s(0), add(s(s(X''''')), X'')) -> RM(s(0), X'')
IFRM(false, s(s(X''''')), add(s(s(Y'''')), X'')) -> RM(s(s(X''''')), X'')
IFRM(false, s(s(X''''')), add(s(0), 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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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(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'')

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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(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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

There are no usable rules for innermost 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) =  1 POL(false) =  0 POL(true) =  0 POL(s(x1)) =  0 POL(RM(x1, x2)) =  x2 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 24`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 2 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 26`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

There are no usable rules for innermost 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) =  1 POL(false) =  0 POL(true) =  0 POL(s(x1)) =  x1 POL(RM(x1, x2)) =  x2 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 29`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 31`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

There are no usable rules for innermost 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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 33`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 27`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

There are no usable rules for innermost 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(0) =  1 POL(false) =  0 POL(true) =  0 POL(s(x1)) =  x1 POL(RM(x1, x2)) =  x2 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 30`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 32`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pair can be strictly oriented:

There are no usable rules for innermost 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(0) =  0 POL(false) =  0 POL(s(x1)) =  1 POL(RM(x1, x2)) =  x2 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 34`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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'')
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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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'')

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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(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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

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:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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:

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

three new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

There are no usable rules for innermost 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(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 25`
`                 ↳Polynomial Ordering`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pair can be strictly oriented:

There are no usable rules for innermost 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(0) =  1 POL(true) =  0 POL(RM(x1, x2)) =  x2 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 7`
`             ↳Nar`
`             ...`
`               →DP Problem 28`
`                 ↳Dependency Graph`
`       →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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Narrowing Transformation`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Narrowing Transformation`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

four new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 36`
`                 ↳Rewriting Transformation`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 37`
`                 ↳Rewriting Transformation`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 38`
`                 ↳Rewriting Transformation`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 39`
`                 ↳Polynomial Ordering`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

ifrm(true, N, add(M, X)) -> rm(N, X)
rm(N, nil) -> nil

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(eq(x1, x2)) =  0 POL(0) =  1 POL(PURGE(x1)) =  1 + x1 POL(false) =  0 POL(ifrm(x1, x2, x3)) =  x3 POL(true) =  0 POL(nil) =  0 POL(rm(x1, x2)) =  x2 POL(s(x1)) =  0 POL(add(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 40`
`                 ↳Dependency Graph`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 41`
`                 ↳Polynomial Ordering`

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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

ifrm(true, N, add(M, X)) -> rm(N, X)
rm(N, nil) -> nil

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 35`
`             ↳Nar`
`             ...`
`               →DP Problem 42`
`                 ↳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
ifrm(true, N, add(M, X)) -> rm(N, X)
purge(nil) -> nil

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes