Termination Proof using AProVE

Term Rewriting System R:
[x]
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A(b(x)) -> B(a(a(x)))
A(b(x)) -> A(a(x))
A(b(x)) -> A(x)
B(c(x)) -> C(b(b(x)))
B(c(x)) -> B(b(x))
B(c(x)) -> B(x)
C(a(x)) -> A(c(c(x)))
C(a(x)) -> C(c(x))
C(a(x)) -> C(x)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
A(b(x)) -> A(x)
A(b(x)) -> A(a(x))
C(a(x)) -> A(c(c(x)))
B(c(x)) -> C(b(b(x)))
A(b(x)) -> B(a(a(x)))

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

A(b(x)) -> B(a(a(x)))

two new Dependency Pairs are created:

A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(u(x''))) -> B(a(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(x)) -> A(a(x))
C(a(x)) -> A(c(c(x)))
B(c(x)) -> C(b(b(x)))
B(c(x)) -> B(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

A(b(x)) -> A(a(x))

two new Dependency Pairs are created:

A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(x''))) -> A(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
C(a(x)) -> C(c(x))
B(c(x)) -> C(b(b(x)))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(x)) -> A(c(c(x)))
C(a(x)) -> C(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

B(c(x)) -> C(b(b(x)))

two new Dependency Pairs are created:

B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(v(x''))) -> C(b(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

A(b(b(x''))) -> A(b(a(a(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

B(c(x)) -> B(b(x))

two new Dependency Pairs are created:

B(c(c(x''))) -> B(c(b(b(x''))))
B(c(v(x''))) -> B(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

A(b(u(x''))) -> A(x'')
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(b(x''))) -> A(b(a(a(x''))))

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

C(a(x)) -> A(c(c(x)))

two new Dependency Pairs are created:

C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(w(x''))) -> A(c(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

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

C(a(x)) -> C(c(x))

two new Dependency Pairs are created:

C(a(a(x''))) -> C(a(c(c(x''))))
C(a(w(x''))) -> C(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pairs:

C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

The following dependency pairs can be strictly oriented:

C(a(w(x''))) -> C(x'')
C(a(w(x''))) -> A(c(x''))

The following usable rules for innermost can be oriented:

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = x₁

POL(C(x₁)) = x₁

POL(B(x₁)) = x₁

POL(v(x₁)) = x₁

POL(b(x₁)) = x₁

POL(a(x₁)) = x₁

POL(w(x₁)) = 1 + x₁

POL(A(x₁)) = x₁

POL(u(x₁)) = x₁

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

C(x₁) -> C(x₁)
A(x₁) -> A(x₁)
a(x₁) -> a(x₁)
c(x₁) -> c(x₁)
w(x₁) -> w(x₁)
B(x₁) -> B(x₁)
v(x₁) -> v(x₁)
b(x₁) -> b(x₁)
u(x₁) -> u(x₁)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

Dependency Pairs:

C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

The following dependency pairs can be strictly oriented:

B(c(v(x''))) -> C(b(x''))
B(c(v(x''))) -> B(x'')

The following usable rules for innermost can be oriented:

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = x₁

POL(C(x₁)) = x₁

POL(B(x₁)) = x₁

POL(v(x₁)) = 1 + x₁

POL(b(x₁)) = x₁

POL(a(x₁)) = x₁

POL(w(x₁)) = 1 + x₁

POL(A(x₁)) = x₁

POL(u(x₁)) = x₁

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

C(x₁) -> C(x₁)
A(x₁) -> A(x₁)
a(x₁) -> a(x₁)
c(x₁) -> c(x₁)
B(x₁) -> B(x₁)
v(x₁) -> v(x₁)
b(x₁) -> b(x₁)
u(x₁) -> u(x₁)
w(x₁) -> w(x₁)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

Dependency Pairs:

C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

The following dependency pairs can be strictly oriented:

A(b(u(x''))) -> A(x'')
A(b(u(x''))) -> B(a(x''))

The following usable rules for innermost can be oriented:

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = x₁

POL(C(x₁)) = x₁

POL(B(x₁)) = x₁

POL(v(x₁)) = 1 + x₁

POL(b(x₁)) = x₁

POL(a(x₁)) = x₁

POL(w(x₁)) = 1 + x₁

POL(A(x₁)) = x₁

POL(u(x₁)) = 1 + x₁

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

C(x₁) -> C(x₁)
A(x₁) -> A(x₁)
a(x₁) -> a(x₁)
c(x₁) -> c(x₁)
B(x₁) -> B(x₁)
b(x₁) -> b(x₁)
u(x₁) -> u(x₁)
w(x₁) -> w(x₁)
v(x₁) -> v(x₁)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

C(a(a(x''))) -> C(a(c(c(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:37 minutes

POL(c(x₁))	= x₁
POL(C(x₁))	= x₁
POL(B(x₁))	= x₁
POL(v(x₁))	= x₁
POL(b(x₁))	= x₁
POL(a(x₁))	= x₁
POL(w(x₁))	= 1 + x₁
POL(A(x₁))	= x₁
POL(u(x₁))	= x₁