Termination Proof using AProVE

Term Rewriting System R:
[a, b, c, d]
f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(f(f(f(j, a), b), c), d) -> F(f(a, b), f(f(a, d), c))
F(f(f(f(j, a), b), c), d) -> F(a, b)
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, a), b), c), d) -> F(a, d)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

F(f(f(f(j, a), b), c), d) -> F(a, d)
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, a), b), c), d) -> F(a, b)
F(f(f(f(j, a), b), c), d) -> F(f(a, b), f(f(a, d), c))

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

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

F(f(f(f(j, a), b), c), d) -> F(f(a, b), f(f(a, d), c))

three new Dependency Pairs are created:

F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(f(f(j, a''), b'''), c''), d), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(f(j, a''), b''), c''), b), f(f(f(a'', b''), f(f(a'', d''), c'')), c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Rewriting Transformation

Dependency Pairs:

F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(f(j, a''), b''), c''), b), f(f(f(a'', b''), f(f(a'', d''), c'')), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(f(f(j, a''), b'''), c''), d), c))
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, a), b), c), d) -> F(a, b)
F(f(f(f(j, a), b), c), d) -> F(a, d)

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

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

F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(f(f(j, a''), b'''), c''), d), c))

one new Dependency Pair is created:

F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(a'', b'''), f(f(a'', d), c'')), c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Rw

...

               →DP Problem 3

                 ↳Rewriting Transformation

Dependency Pairs:

F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(a'', b'''), f(f(a'', d), c'')), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, a), b), c), d) -> F(a, d)
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, a), b), c), d) -> F(a, b)
F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(f(j, a''), b''), c''), b), f(f(f(a'', b''), f(f(a'', d''), c'')), c))

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

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

F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(f(j, a''), b''), c''), b), f(f(f(a'', b''), f(f(a'', d''), c'')), c))

one new Dependency Pair is created:

F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(a'', b''), f(f(a'', b), c'')), f(f(f(a'', b''), f(f(a'', d''), c'')), c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Rw

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(a'', b''), f(f(a'', b), c'')), f(f(f(a'', b''), f(f(a'', d''), c'')), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, a), b), c), d) -> F(a, d)
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, a), b), c), d) -> F(a, b)
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(a'', b'''), f(f(a'', d), c'')), c))

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

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

F(f(f(f(j, a), b), c), d) -> F(a, b)

four new Dependency Pairs are created:

F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(f(j, a''), b'''), c''), b'')
F(f(f(f(j, f(f(f(j, f(f(j, a''''), b'''')), b'0), c'''')), b''), c), d) -> F(f(f(f(j, f(f(j, a''''), b'''')), b'0), c''''), b'')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c'')), b'), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c''), b')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c'')), b''), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c''), b'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Rw

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c'')), b''), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c''), b'')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c'')), b'), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c''), b')
F(f(f(f(j, f(f(f(j, f(f(j, a''''), b'''')), b'0), c'''')), b''), c), d) -> F(f(f(f(j, f(f(j, a''''), b'''')), b'0), c''''), b'')
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(f(j, a''), b'''), c''), b'')
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(a'', b'''), f(f(a'', d), c'')), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, a), b), c), d) -> F(a, d)
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(a'', b''), f(f(a'', b), c'')), f(f(f(a'', b''), f(f(a'', d''), c'')), c))

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

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

F(f(f(f(j, a), b), c), d) -> F(a, d)

seven new Dependency Pairs are created:

F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(j, a''), b''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(j, a''''), b'''')), b''), c'''')), b), c), d') -> F(f(f(f(j, f(f(j, a''''), b'''')), b''), c''''), d')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b''), c'')), b), c), d') -> F(f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b''), c''), d')
F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(j, a''''''), b'''''')), b'0''), c'''''')), b''''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(j, a''''''), b'''''')), b'0''), c'''''')), b''''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b'''''''), c'''''')), b''0'''), c'''')), b'''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b'''''''), c'''''')), b''0'''), c'''')), b'''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b''''''), c'''''')), b'0''), c'''')), b''''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b''''''), c'''''')), b'0''), c'''')), b''''), c''), d'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Rw

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b''''''), c'''''')), b'0''), c'''')), b''''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b''''''), c'''''')), b'0''), c'''')), b''''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b'''''''), c'''''')), b''0'''), c'''')), b'''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(f(j, a''''''), b'''''''), c'''''')), b''0'''), c'''')), b'''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, f(f(j, a''''''), b'''''')), b'0''), c'''''')), b''''), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, f(f(j, a''''''), b'''''')), b'0''), c'''''')), b''''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b''), c'')), b), c), d') -> F(f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b''), c''), d')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c'')), b), c), d'') -> F(f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(j, a''''), b'''')), b''), c'''')), b), c), d') -> F(f(f(f(j, f(f(j, a''''), b'''')), b''), c''''), d')
F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(f(j, a''), b''), c''), d'')
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c'')), b'), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b'''''), c'''')), b''0'), c''), b')
F(f(f(f(j, f(f(f(j, f(f(j, a''''), b'''')), b'0), c'''')), b''), c), d) -> F(f(f(f(j, f(f(j, a''''), b'''')), b'0), c''''), b'')
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(f(j, a''), b'''), c''), b'')
F(f(f(f(j, f(f(f(j, a''), b''), c'')), b), c), d'') -> F(f(f(a'', b''), f(f(a'', b), c'')), f(f(f(a'', b''), f(f(a'', d''), c'')), c))
F(f(f(f(j, f(f(f(j, a''), b'''), c'')), b''), c), d) -> F(f(f(a'', b'''), f(f(a'', b''), c'')), f(f(f(a'', b'''), f(f(a'', d), c'')), c))
F(f(f(f(j, f(f(j, a''), b'')), b), c''), d'') -> F(f(f(f(j, a''), b''), b), f(f(a'', b''), f(f(a'', c''), d'')))
F(f(f(f(j, a), b), c), d) -> F(f(a, d), c)
F(f(f(f(j, f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c'')), b''), c), d) -> F(f(f(f(j, f(f(f(j, a''''), b''''), c'''')), b'0), c''), b'')

Rule:

f(f(f(f(j, a), b), c), d) -> f(f(a, b), f(f(a, d), c))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:06 minutes