Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 1

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

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

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 1

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

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

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

F(0, s(s(0)), s(s(0))) -> F(0, 0, s(s(0)))
F(0, s(s(s(0))), s(s(0))) -> F(0, s(0), s(s(0)))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(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 11

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

seven new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> x₂
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Nar

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the 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)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Nar

...

               →DP Problem 18

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

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

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 4

         ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

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

six new Dependency Pairs are created:

F(s(0), s(0), s(0)) -> F(0, 0, s(s(0)))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(0), s(s(0)), s(0)) -> F(0, s(0), s(s(0)))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(0), s(s(s(y'))), s(0)) -> F(0, s(s(y')), f(0, y', s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 20

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

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

five new Dependency Pairs are created:

F(s(s(x'')), s(0), 0) -> F(s(x''), 0, s(0))
F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 22

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

seven new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 23

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(s(s(s(s(x'''''')))), s(0), 0) -> F(s(s(s(x''''''))), 0, s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 25

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

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

one new Dependency Pair is created:

F(s(s(s(s(s(x''''''''))))), 0, s(0)) -> F(s(s(s(s(x'''''''')))), s(0), 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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 27

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the 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)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 32

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 24

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

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

11 new Dependency Pairs are created:

F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 26

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

seven new Dependency Pairs are created:

F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 28

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

11 new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 29

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

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

17 new Dependency Pairs are created:

F(s(s(0)), s(y'''), s(s(0))) -> F(s(s(0)), s(y'''), s(0))
F(s(s(x'''')), s(0), s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(x'''')), s(s(y'''')), s(s(0))) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'''), s(y'''), s(s(s(0)))) -> F(s(x'''), s(y'''), s(s(0)))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(s(0)), s(y''''), s(s(s(0)))) -> F(s(s(0)), s(y''''), s(s(0)))
F(s(s(x''''')), s(0), s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(x''''')), s(s(y''''')), s(s(s(0)))) -> F(s(s(x''''')), s(s(y''''')), s(s(0)))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(s(x'''''')), s(s(y'''''')), s(s(0))) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(0))), s(s(y''''''''')), s(s(0))) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x'''''''))), s(s(0)), s(s(0))) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(s(0))) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 30

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

eight new Dependency Pairs are created:

F(s(s(s(0))), s(s(y''''')), s(0)) -> F(s(s(s(0))), s(s(y''''')), 0)
F(s(s(s(x''''''))), s(s(0)), s(0)) -> F(s(s(s(x''''''))), s(s(0)), 0)
F(s(s(s(x''''''))), s(s(s(y''''''))), s(0)) -> F(s(s(s(x''''''))), s(s(s(y''''''))), 0)
F(s(s(s(0))), s(s(y'''''')), s(0)) -> F(s(s(s(0))), s(s(y'''''')), 0)
F(s(s(s(x''''''''))), s(s(s(y''''''''))), s(0)) -> F(s(s(s(x''''''''))), s(s(s(y''''''''))), 0)
F(s(s(s(s(0)))), s(s(s(y'''''''''''))), s(0)) -> F(s(s(s(s(0)))), s(s(s(y'''''''''''))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(0))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(0))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(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 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 31

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> x₁
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 33

                 ↳Dependency Graph

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 34

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

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

seven new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 35

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the 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)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 36

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:57 minutes

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