p(

p(

p(

R

↳Dependency Pair Analysis

P(m,n, s(r)) -> P(m,r,n)

P(m, s(n), 0) -> P(0,n,m)

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

**P( m, s(n), 0) -> P(0, n, m)**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

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:

P(m,n, s(r)) -> P(m,r,n)

P(m'', s(r''), s(r0)) -> P(m'',r0, s(r''))

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

**P( m'', 0, s(s(n''))) -> P(m'', s(n''), 0)**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

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:

P(m, s(n), 0) -> P(0,n,m)

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)

P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))

P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))

The transformation is resulting in two new DP problems:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Forward Instantiation Transformation

**P(0, s(s( n'')), 0) -> P(0, s(n''), 0)**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 5

↳Polynomial Ordering

**P(0, s(s(s( n''''))), 0) -> P(0, s(s(n'''')), 0)**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

The following dependency pair can be strictly oriented:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

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(P(x)_{1}, x_{2}, x_{3})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 9

↳Dependency Graph

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Forward Instantiation Transformation

**P(s(s( n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

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:

P(m'', s(r''), s(r0)) -> P(m'',r0, s(r''))

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))

P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 6

↳Forward Instantiation Transformation

**P( m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

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:

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)

P(s(s(n'''''')), 0, s(s(0))) -> P(s(s(n'''''')), s(0), 0)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 7

↳Forward Instantiation Transformation

**P( m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

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:

P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))

P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 8

↳Polynomial Ordering

**P(s(0), s(s(s( n''''''))), 0) -> P(0, s(s(n'''''')), s(0))**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

The following dependency pairs can be strictly oriented:

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))

P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))

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(P(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 10

↳Dependency Graph

**P(s( r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

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

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 11

↳Polynomial Ordering

**P( m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))**

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

The following dependency pair can be strictly oriented:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))

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(P(x)_{1}, x_{2}, x_{3})= 1 + x _{1}+ x_{2}+ x_{3}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 12

↳Dependency Graph

p(m,n, s(r)) -> p(m,r,n)

p(m, s(n), 0) -> p(0,n,m)

p(m, 0, 0) ->m

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes