Termination Proof using AProVE

Term Rewriting System R:
[x, y, u, z]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
<='(s(x), s(y)) -> <='(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) -> -'(z, s(x))
F(s(x), s(y), z, u) -> IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> <='(x, y)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) -> -'(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

-'(s(x), s(y)) -> -'(x, y)

one new Dependency Pair is created:

-'(s(s(x'')), s(s(y''))) -> -'(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

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(y''))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(y''))

one new Dependency Pair is created:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(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

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(s(s(x'''')), s(s(y'''')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pair can be strictly oriented:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(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(-'(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

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

<='(s(x), s(y)) -> <='(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

<='(s(x), s(y)) -> <='(x, y)

one new Dependency Pair is created:

<='(s(s(x'')), s(s(y''))) -> <='(s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

<='(s(s(x'')), s(s(y''))) -> <='(s(x''), s(y''))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

<='(s(s(x'')), s(s(y''))) -> <='(s(x''), s(y''))

one new Dependency Pair is created:

<='(s(s(s(x''''))), s(s(s(y'''')))) -> <='(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

<='(s(s(s(x''''))), s(s(s(y'''')))) -> <='(s(s(x'''')), s(s(y'''')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pair can be strictly oriented:

<='(s(s(s(x''''))), s(s(s(y'''')))) -> <='(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(s(x₁)) = 1 + x₁

POL(<='(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)

two new Dependency Pairs are created:

F(s(0), s(y'), z, u) -> F(s(0), y', z, u)
F(s(s(y'')), s(s(x'')), z, u) -> F(s(s(y'')), -(x'', y''), z, u)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pairs:

F(s(s(y'')), s(s(x'')), z, u) -> F(s(s(y'')), -(x'', y''), z, u)
F(s(0), s(y'), z, u) -> F(s(0), y', z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(s(y'')), s(s(x'')), z, u) -> F(s(s(y'')), -(x'', y''), z, u)

two new Dependency Pairs are created:

F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Instantiation Transformation

Dependency Pairs:

F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(0), s(y'), z, u) -> F(s(0), y', z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)

five new Dependency Pairs are created:

F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(0), 0, z'', u'') -> F(0, u'', -(z'', s(0)), u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Instantiation Transformation

Dependency Pairs:

F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(0), s(y'), z, u) -> F(s(0), y', z, u)
F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(x), s(y), z, u) -> F(x, u, z, u)

six new Dependency Pairs are created:

F(s(x''), s(y'), z'', s(y')) -> F(x'', s(y'), z'', s(y'))
F(s(0), s(y'), z'', u'') -> F(0, u'', z'', u'')
F(s(s(0)), s(y'), z'', u'') -> F(s(0), u'', z'', u'')
F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')
F(s(0), s(y'), z', s(y')) -> F(0, s(y'), z', s(y'))
F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))
F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')
F(s(s(0)), s(y'), z'', u'') -> F(s(0), u'', z'', u'')
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(x''), s(y'), z'', s(y')) -> F(x'', s(y'), z'', s(y'))
F(s(0), s(y'), z, u) -> F(s(0), y', z, u)
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(0), s(y'), z, u) -> F(s(0), y', z, u)

three new Dependency Pairs are created:

F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(s(0)), s(y'), z'', u'') -> F(s(0), u'', z'', u'')
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(x''), s(y'), z'', s(y')) -> F(x'', s(y'), z'', s(y'))
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(x''), s(y'), z'', s(y')) -> F(x'', s(y'), z'', s(y'))

eight new Dependency Pairs are created:

F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(0)), s(y'), z'', u'') -> F(s(0), u'', z'', u'')
F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(s(0)), s(y'), z'', u'') -> F(s(0), u'', z'', u'')

three new Dependency Pairs are created:

F(s(s(0)), s(y'), z'''', 0) -> F(s(0), 0, z'''', 0)
F(s(s(0)), s(y'), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(y'), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(0)), s(y'), z'''', 0) -> F(s(0), 0, z'''', 0)
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(0)), s(y'), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(y'), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))
F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(s(s(y'''))), s(y'), z'', u'') -> F(s(s(y''')), u'', z'', u'')

17 new Dependency Pairs are created:

F(s(s(s(0))), s(y'), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(s(y'''')))), s(y'), z''', s(s(s(x''')))) -> F(s(s(s(y''''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(y''''))), s(y'), z'''', 0) -> F(s(s(y'''')), 0, z'''', 0)
F(s(s(s(0))), s(y'), z'''', 0) -> F(s(s(0)), 0, z'''', 0)
F(s(s(s(s(y''''')))), s(y'), z'''', 0) -> F(s(s(s(y'''''))), 0, z'''', 0)
F(s(s(s(s(y''''')))), s(y'), z'''', s(y''0)) -> F(s(s(s(y'''''))), s(y''0), z'''', s(y''0))
F(s(s(s(y''''))), s(y'), z'''', s(y''0)) -> F(s(s(y'''')), s(y''0), z'''', s(y''0))
F(s(s(s(s(0)))), s(y'), z''', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z''', s(s(x'''''''')))
F(s(s(s(s(s(y'''''))))), s(y'), z''', s(s(s(x''''''')))) -> F(s(s(s(s(y''''')))), s(s(s(x'''''''))), z''', s(s(s(x'''''''))))
F(s(s(s(y''''))), s(y'), z''', s(y'0''')) -> F(s(s(y'''')), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(0)))), s(y'), z''', s(y'''''')) -> F(s(s(s(0))), s(y''''''), z''', s(y''''''))
F(s(s(s(s(s(y'''''''))))), s(y'), z''', s(y''''')) -> F(s(s(s(s(y''''''')))), s(y'''''), z''', s(y'''''))
F(s(s(s(s(y''''''''')))), s(y'), z''', s(y'0''')) -> F(s(s(s(y'''''''''))), s(y'0'''), z''', s(y'0'''))
F(s(s(s(0))), s(y'), z''', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z''', s(s(y'''''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z''', s(s(y''0''''')))
F(s(s(s(0))), s(y'), z''', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z''', s(s(y''''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z''', s(s(y''0'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(s(y''''')))), s(y'), z'''', 0) -> F(s(s(s(y'''''))), 0, z'''', 0)
F(s(s(s(0))), s(y'), z'''', 0) -> F(s(s(0)), 0, z'''', 0)
F(s(s(s(y''''))), s(y'), z'''', 0) -> F(s(s(y'''')), 0, z'''', 0)
F(s(s(s(s(y''''''''')))), s(y'), z''', s(y'0''')) -> F(s(s(s(y'''''''''))), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(s(y'''''''))))), s(y'), z''', s(y''''')) -> F(s(s(s(s(y''''''')))), s(y'''''), z''', s(y'''''))
F(s(s(s(0))), s(y'), z''', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z''', s(s(y''0'''')))
F(s(s(s(0))), s(y'), z''', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z''', s(s(y''''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z''', s(s(y''0''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z''', s(s(y'''''''')))
F(s(s(s(s(0)))), s(y'), z''', s(y'''''')) -> F(s(s(s(0))), s(y''''''), z''', s(y''''''))
F(s(s(s(s(s(y'''''))))), s(y'), z''', s(s(s(x''''''')))) -> F(s(s(s(s(y''''')))), s(s(s(x'''''''))), z''', s(s(s(x'''''''))))
F(s(s(s(y''''))), s(y'), z''', s(y'0''')) -> F(s(s(y'''')), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(0)))), s(y'), z''', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z''', s(s(x'''''''')))
F(s(s(s(y''''))), s(y'), z'''', s(y''0)) -> F(s(s(y'''')), s(y''0), z'''', s(y''0))
F(s(s(s(s(y''''')))), s(y'), z'''', s(y''0)) -> F(s(s(s(y'''''))), s(y''0), z'''', s(y''0))
F(s(s(s(s(y'''')))), s(y'), z''', s(s(s(x''')))) -> F(s(s(s(y''''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(0))), s(y'), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(0)), s(y'), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(y'), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(0)), s(y'), z'''', 0) -> F(s(0), 0, z'''', 0)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

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

F(s(s(y''''')), s(y'), z', s(y')) -> F(s(y'''''), s(y'), z', s(y'))

28 new Dependency Pairs are created:

F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(s(y''')))), s(s(s(x'''))), z''', s(s(s(x''')))) -> F(s(s(s(y'''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(y'''''''))), s(y'0), z''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z''', s(y'0))
F(s(s(0)), s(s(y'''''')), z'', s(s(y''''''))) -> F(s(0), s(s(y'''''')), z'', s(s(y'''''')))
F(s(s(0)), s(s(y''0'')), z''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z''', s(s(y''0'')))
F(s(s(s(s(0)))), s(s(x'''''''')), z'', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z'', s(s(x'''''''')))
F(s(s(s(s(s(y''''''))))), s(s(s(x'''''''))), z'', s(s(s(x''''''')))) -> F(s(s(s(s(y'''''')))), s(s(s(x'''''''))), z'', s(s(s(x'''''''))))
F(s(s(s(x''''''))), s(y''), z'', s(y'')) -> F(s(s(x'''''')), s(y''), z'', s(y''))
F(s(s(s(s(0)))), s(y''), z'', s(y'')) -> F(s(s(s(0))), s(y''), z'', s(y''))
F(s(s(s(s(s(y'''''''))))), s(y''), z'', s(y'')) -> F(s(s(s(s(y''''''')))), s(y''), z'', s(y''))
F(s(s(s(s(y''''''''')))), s(y''), z'', s(y'')) -> F(s(s(s(y'''''''''))), s(y''), z'', s(y''))
F(s(s(s(0))), s(s(y'''''''')), z'', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z'', s(s(y'''''''')))
F(s(s(s(0))), s(s(y''0''''')), z'', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z'', s(s(y''0''''')))
F(s(s(s(0))), s(s(y''''''')), z'', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z'', s(s(y''''''')))
F(s(s(s(0))), s(s(y''0'''')), z'', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z'', s(s(y''0'''')))
F(s(s(s(s(0)))), s(s(x''''''')), z'', s(s(x'''''''))) -> F(s(s(s(0))), s(s(x''''''')), z'', s(s(x''''''')))
F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''))), z'', s(s(s(x''''')))) -> F(s(s(s(s(y''''''')))), s(s(s(x'''''))), z'', s(s(s(x'''''))))
F(s(s(s(s(s(y'''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(y''''''')))), s(y'''), z'', s(y''')) -> F(s(s(s(y'''''''))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(0))))), s(s(x'''''''''')), z'', s(s(x''''''''''))) -> F(s(s(s(s(0)))), s(s(x'''''''''')), z'', s(s(x'''''''''')))
F(s(s(s(s(s(s(y''''''')))))), s(s(s(x'''''''''))), z'', s(s(s(x''''''''')))) -> F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''''''))), z'', s(s(s(x'''''''''))))
F(s(s(s(s(s(0))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(0)))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(s(y''''''''')))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(s(y'''''''''))))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(y'''''''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(0)))), s(s(y'''''''''')), z'', s(s(y''''''''''))) -> F(s(s(s(0))), s(s(y'''''''''')), z'', s(s(y'''''''''')))
F(s(s(s(s(0)))), s(s(y''0''''''')), z'', s(s(y''0'''''''))) -> F(s(s(s(0))), s(s(y''0''''''')), z'', s(s(y''0''''''')))
F(s(s(s(s(0)))), s(s(y''''''''')), z'', s(s(y'''''''''))) -> F(s(s(s(0))), s(s(y''''''''')), z'', s(s(y''''''''')))
F(s(s(s(s(0)))), s(s(y''0'''''')), z'', s(s(y''0''''''))) -> F(s(s(s(0))), s(s(y''0'''''')), z'', s(s(y''0'''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 18

                 ↳Polynomial Ordering

Dependency Pairs:

F(s(s(0)), s(y'), z'''', 0) -> F(s(0), 0, z'''', 0)
F(s(s(s(0))), s(y'), z'''', 0) -> F(s(s(0)), 0, z'''', 0)
F(s(s(s(y''''))), s(y'), z'''', 0) -> F(s(s(y'''')), 0, z'''', 0)
F(s(s(s(s(s(y'''''''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(s(y''''''''')))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(s(y'''''''''))))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(0))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(0)))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(s(y''''''')))))), s(s(s(x'''''''''))), z'', s(s(s(x''''''''')))) -> F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''''''))), z'', s(s(s(x'''''''''))))
F(s(s(s(s(0)))), s(s(y''0'''''')), z'', s(s(y''0''''''))) -> F(s(s(s(0))), s(s(y''0'''''')), z'', s(s(y''0'''''')))
F(s(s(s(s(0)))), s(s(y''''''''')), z'', s(s(y'''''''''))) -> F(s(s(s(0))), s(s(y''''''''')), z'', s(s(y''''''''')))
F(s(s(s(s(0)))), s(s(y''0''''''')), z'', s(s(y''0'''''''))) -> F(s(s(s(0))), s(s(y''0''''''')), z'', s(s(y''0''''''')))
F(s(s(s(s(0)))), s(s(y'''''''''')), z'', s(s(y''''''''''))) -> F(s(s(s(0))), s(s(y'''''''''')), z'', s(s(y'''''''''')))
F(s(s(s(s(s(0))))), s(s(x'''''''''')), z'', s(s(x''''''''''))) -> F(s(s(s(s(0)))), s(s(x'''''''''')), z'', s(s(x'''''''''')))
F(s(s(s(s(y''''''')))), s(y'''), z'', s(y''')) -> F(s(s(s(y'''''''))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(y'''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''))), z'', s(s(s(x''''')))) -> F(s(s(s(s(y''''''')))), s(s(s(x'''''))), z'', s(s(s(x'''''))))
F(s(s(s(s(0)))), s(s(x''''''')), z'', s(s(x'''''''))) -> F(s(s(s(0))), s(s(x''''''')), z'', s(s(x''''''')))
F(s(s(s(s(y''''''''')))), s(y''), z'', s(y'')) -> F(s(s(s(y'''''''''))), s(y''), z'', s(y''))
F(s(s(s(s(s(y'''''''))))), s(y''), z'', s(y'')) -> F(s(s(s(s(y''''''')))), s(y''), z'', s(y''))
F(s(s(s(0))), s(s(y''0'''')), z'', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z'', s(s(y''0'''')))
F(s(s(s(0))), s(s(y''''''')), z'', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z'', s(s(y''''''')))
F(s(s(s(0))), s(s(y''0''''')), z'', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z'', s(s(y''0''''')))
F(s(s(s(0))), s(s(y'''''''')), z'', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z'', s(s(y'''''''')))
F(s(s(s(s(0)))), s(y''), z'', s(y'')) -> F(s(s(s(0))), s(y''), z'', s(y''))
F(s(s(s(s(s(y''''''))))), s(s(s(x'''''''))), z'', s(s(s(x''''''')))) -> F(s(s(s(s(y'''''')))), s(s(s(x'''''''))), z'', s(s(s(x'''''''))))
F(s(s(s(x''''''))), s(y''), z'', s(y'')) -> F(s(s(x'''''')), s(y''), z'', s(y''))
F(s(s(s(s(0)))), s(s(x'''''''')), z'', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z'', s(s(x'''''''')))
F(s(s(s(s(y''')))), s(s(s(x'''))), z''', s(s(s(x''')))) -> F(s(s(s(y'''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(s(y''''''''')))), s(y'), z''', s(y'0''')) -> F(s(s(s(y'''''''''))), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(s(y'''''''))))), s(y'), z''', s(y''''')) -> F(s(s(s(s(y''''''')))), s(y'''''), z''', s(y'''''))
F(s(s(s(y'''''''))), s(y'0), z''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z''', s(y'0))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z''', s(s(y''0'''')))
F(s(s(s(0))), s(y'), z''', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z''', s(s(y''''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z''', s(s(y''0''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z''', s(s(y'''''''')))
F(s(s(s(s(0)))), s(y'), z''', s(y'''''')) -> F(s(s(s(0))), s(y''''''), z''', s(y''''''))
F(s(s(s(s(s(y'''''))))), s(y'), z''', s(s(s(x''''''')))) -> F(s(s(s(s(y''''')))), s(s(s(x'''''''))), z''', s(s(s(x'''''''))))
F(s(s(s(y''''))), s(y'), z''', s(y'0''')) -> F(s(s(y'''')), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(0)))), s(y'), z''', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z''', s(s(x'''''''')))
F(s(s(s(y''''))), s(y'), z'''', s(y''0)) -> F(s(s(y'''')), s(y''0), z'''', s(y''0))
F(s(s(s(s(y''''')))), s(y'), z'''', s(y''0)) -> F(s(s(s(y'''''))), s(y''0), z'''', s(y''0))
F(s(s(s(s(y'''')))), s(y'), z''', s(s(s(x''')))) -> F(s(s(s(y''''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(0))), s(y'), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(0)), s(s(y''0'')), z''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z''', s(s(y''0'')))
F(s(s(0)), s(s(y'''''')), z'', s(s(y''''''))) -> F(s(0), s(s(y'''''')), z'', s(s(y'''''')))
F(s(s(0)), s(y'), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(y'), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(s(s(y''''')))), s(y'), z'''', 0) -> F(s(s(s(y'''''))), 0, z'''', 0)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(s(0)), s(y'), z'''', 0) -> F(s(0), 0, z'''', 0)
F(s(s(s(0))), s(y'), z'''', 0) -> F(s(s(0)), 0, z'''', 0)
F(s(s(s(y''''))), s(y'), z'''', 0) -> F(s(s(y'''')), 0, z'''', 0)
F(s(s(s(s(s(y'''''''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(s(y''''''''')))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(s(y'''''''''))))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(0))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(0)))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(s(y''''''')))))), s(s(s(x'''''''''))), z'', s(s(s(x''''''''')))) -> F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''''''))), z'', s(s(s(x'''''''''))))
F(s(s(s(s(0)))), s(s(y''0'''''')), z'', s(s(y''0''''''))) -> F(s(s(s(0))), s(s(y''0'''''')), z'', s(s(y''0'''''')))
F(s(s(s(s(0)))), s(s(y''''''''')), z'', s(s(y'''''''''))) -> F(s(s(s(0))), s(s(y''''''''')), z'', s(s(y''''''''')))
F(s(s(s(s(0)))), s(s(y''0''''''')), z'', s(s(y''0'''''''))) -> F(s(s(s(0))), s(s(y''0''''''')), z'', s(s(y''0''''''')))
F(s(s(s(s(0)))), s(s(y'''''''''')), z'', s(s(y''''''''''))) -> F(s(s(s(0))), s(s(y'''''''''')), z'', s(s(y'''''''''')))
F(s(s(s(s(s(0))))), s(s(x'''''''''')), z'', s(s(x''''''''''))) -> F(s(s(s(s(0)))), s(s(x'''''''''')), z'', s(s(x'''''''''')))
F(s(s(s(s(y''''''')))), s(y'''), z'', s(y''')) -> F(s(s(s(y'''''''))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(y'''''''))))), s(y'''), z'', s(y''')) -> F(s(s(s(s(y''''''')))), s(y'''), z'', s(y'''))
F(s(s(s(s(s(y'''''''))))), s(s(s(x'''''))), z'', s(s(s(x''''')))) -> F(s(s(s(s(y''''''')))), s(s(s(x'''''))), z'', s(s(s(x'''''))))
F(s(s(s(s(0)))), s(s(x''''''')), z'', s(s(x'''''''))) -> F(s(s(s(0))), s(s(x''''''')), z'', s(s(x''''''')))
F(s(s(s(s(y''''''''')))), s(y''), z'', s(y'')) -> F(s(s(s(y'''''''''))), s(y''), z'', s(y''))
F(s(s(s(s(s(y'''''''))))), s(y''), z'', s(y'')) -> F(s(s(s(s(y''''''')))), s(y''), z'', s(y''))
F(s(s(s(0))), s(s(y''0'''')), z'', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z'', s(s(y''0'''')))
F(s(s(s(0))), s(s(y''''''')), z'', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z'', s(s(y''''''')))
F(s(s(s(0))), s(s(y''0''''')), z'', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z'', s(s(y''0''''')))
F(s(s(s(0))), s(s(y'''''''')), z'', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z'', s(s(y'''''''')))
F(s(s(s(s(0)))), s(y''), z'', s(y'')) -> F(s(s(s(0))), s(y''), z'', s(y''))
F(s(s(s(s(s(y''''''))))), s(s(s(x'''''''))), z'', s(s(s(x''''''')))) -> F(s(s(s(s(y'''''')))), s(s(s(x'''''''))), z'', s(s(s(x'''''''))))
F(s(s(s(x''''''))), s(y''), z'', s(y'')) -> F(s(s(x'''''')), s(y''), z'', s(y''))
F(s(s(s(s(0)))), s(s(x'''''''')), z'', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z'', s(s(x'''''''')))
F(s(s(s(s(y''')))), s(s(s(x'''))), z''', s(s(s(x''')))) -> F(s(s(s(y'''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(s(y''''''''')))), s(y'), z''', s(y'0''')) -> F(s(s(s(y'''''''''))), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(s(y'''''''))))), s(y'), z''', s(y''''')) -> F(s(s(s(s(y''''''')))), s(y'''''), z''', s(y'''''))
F(s(s(s(y'''''''))), s(y'0), z''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z''', s(y'0))
F(s(s(s(0))), s(s(x''''')), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0''''))) -> F(s(s(0)), s(s(y''0'''')), z''', s(s(y''0'''')))
F(s(s(s(0))), s(y'), z''', s(s(y'''''''))) -> F(s(s(0)), s(s(y''''''')), z''', s(s(y''''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''0'''''))) -> F(s(s(0)), s(s(y''0''''')), z''', s(s(y''0''''')))
F(s(s(s(0))), s(y'), z''', s(s(y''''''''))) -> F(s(s(0)), s(s(y'''''''')), z''', s(s(y'''''''')))
F(s(s(s(s(0)))), s(y'), z''', s(y'''''')) -> F(s(s(s(0))), s(y''''''), z''', s(y''''''))
F(s(s(s(s(s(y'''''))))), s(y'), z''', s(s(s(x''''''')))) -> F(s(s(s(s(y''''')))), s(s(s(x'''''''))), z''', s(s(s(x'''''''))))
F(s(s(s(y''''))), s(y'), z''', s(y'0''')) -> F(s(s(y'''')), s(y'0'''), z''', s(y'0'''))
F(s(s(s(s(0)))), s(y'), z''', s(s(x''''''''))) -> F(s(s(s(0))), s(s(x'''''''')), z''', s(s(x'''''''')))
F(s(s(s(y''''))), s(y'), z'''', s(y''0)) -> F(s(s(y'''')), s(y''0), z'''', s(y''0))
F(s(s(s(s(y''''')))), s(y'), z'''', s(y''0)) -> F(s(s(s(y'''''))), s(y''0), z'''', s(y''0))
F(s(s(s(s(y'''')))), s(y'), z''', s(s(s(x''')))) -> F(s(s(s(y''''))), s(s(s(x'''))), z''', s(s(s(x'''))))
F(s(s(s(0))), s(y'), z''', s(s(x'''''))) -> F(s(s(0)), s(s(x''''')), z''', s(s(x''''')))
F(s(s(s(y'''''''))), s(y'0), z'''', s(y'0)) -> F(s(s(y''''''')), s(y'0), z'''', s(y'0))
F(s(s(s(s(y''''')))), s(y'''), z'''', s(y''')) -> F(s(s(s(y'''''))), s(y'''), z'''', s(y'''))
F(s(s(s(0))), s(y'''), z'''', s(y''')) -> F(s(s(0)), s(y'''), z'''', s(y'''))
F(s(s(s(s(y''')))), s(s(s(x''''))), z''', s(s(s(x'''')))) -> F(s(s(s(y'''))), s(s(s(x''''))), z''', s(s(s(x''''))))
F(s(s(0)), s(s(y''0'')), z''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z''', s(s(y''0'')))
F(s(s(0)), s(s(y'''''')), z'', s(s(y''''''))) -> F(s(0), s(s(y'''''')), z'', s(s(y'''''')))
F(s(s(0)), s(y'), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(y'), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(0)), s(s(y''0'')), z'''', s(s(y''0''))) -> F(s(0), s(s(y''0'')), z'''', s(s(y''0'')))
F(s(s(0)), s(s(y''''')), z'''', s(s(y'''''))) -> F(s(0), s(s(y''''')), z'''', s(s(y''''')))
F(s(s(x'''')), s(y'0), z'''', s(y'0)) -> F(s(x''''), s(y'0), z'''', s(y'0))
F(s(s(s(y'''))), 0, z'', u'') -> F(s(s(y''')), u'', -(z'', s(s(s(y''')))), u'')
F(s(s(0)), 0, z'', u'') -> F(s(0), u'', -(z'', s(s(0))), u'')
F(s(x''), 0, z'', 0) -> F(x'', 0, -(z'', s(x'')), 0)
F(s(x'), 0, z', 0) -> F(x', 0, -(z', s(x')), 0)
F(s(s(s(s(y''''')))), s(y'), z'''', 0) -> F(s(s(s(y'''''))), 0, z'''', 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(0) = 0

POL(s(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = 0

POL(F(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₄

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 19

                 ↳Dependency Graph

Dependency Pairs:

F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)
F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))
F(s(0), s(0), z', 0) -> F(s(0), 0, z', 0)
F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 3 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 20

                 ↳Polynomial Ordering

Dependency Pair:

F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(s(s(s(y'))), s(s(s(x'))), z, u) -> F(s(s(s(y'))), -(x', y'), z, u)

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

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 1

POL(s(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = x₁

POL(F(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 21

                 ↳Polynomial Ordering

Dependency Pair:

F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(s(s(0)), s(s(x''')), z, u) -> F(s(s(0)), x''', z, u)

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(0) = 0

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃, x₄)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

Dependency Pairs:

F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', s(y''''))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(0), s(s(y''')), z'', u'') -> F(s(0), s(y'''), z'', u'')
F(s(0), s(s(y''0)), z', s(y'''')) -> F(s(0), s(y''0), z', 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(0) = 0

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃, x₄)) = 1 + x₂

resulting in one new DP problem.

Innermost Termination of R successfully shown.
Duration:
0:53 minutes

POL(s(x₁))	= 1 + x₁
POL(<='(x₁, x₂))	= 1 + x₁

POL(0)	= 0
POL(s(x₁))	= 1 + x₁
POL(-(x₁, x₂))	= 0
POL(F(x₁, x₂, x₃, x₄))	= 1 + x₁ + x₄