Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

V(a(a(x))) -> U(v(x))
V(a(a(x))) -> V(x)
V(a(c(x))) -> U(b(d(x)))
W(a(a(x))) -> U(w(x))
W(a(a(x))) -> W(x)
W(a(c(x))) -> U(b(d(x)))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

V(a(a(x))) -> V(x)

Rules:

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

Strategy:

innermost

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

V(a(a(x))) -> V(x)

one new Dependency Pair is created:

V(a(a(a(a(x''))))) -> V(a(a(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

V(a(a(a(a(x''))))) -> V(a(a(x'')))

Rules:

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

Strategy:

innermost

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

V(a(a(a(a(x''))))) -> V(a(a(x'')))

one new Dependency Pair is created:

V(a(a(a(a(a(a(x''''))))))) -> V(a(a(a(a(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

V(a(a(a(a(a(a(x''''))))))) -> V(a(a(a(a(x'''')))))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

V(a(a(a(a(a(a(x''''))))))) -> V(a(a(a(a(x'''')))))

The following usable rule for innermost can be oriented:

a(c(d(x))) -> c(x)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = x₁

POL(V(x₁)) = 1 + x₁

POL(d(x₁)) = x₁

POL(a(x₁)) = 1 + x₁

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

V(x₁) -> V(x₁)
a(x₁) -> a(x₁)
c(x₁) -> c(x₁)
d(x₁) -> d(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pair:

W(a(a(x))) -> W(x)

Rules:

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

Strategy:

innermost

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

W(a(a(x))) -> W(x)

one new Dependency Pair is created:

W(a(a(a(a(x''))))) -> W(a(a(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pair:

W(a(a(a(a(x''))))) -> W(a(a(x'')))

Rules:

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

Strategy:

innermost

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

W(a(a(a(a(x''))))) -> W(a(a(x'')))

one new Dependency Pair is created:

W(a(a(a(a(a(a(x''''))))))) -> W(a(a(a(a(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pair:

W(a(a(a(a(a(a(x''''))))))) -> W(a(a(a(a(x'''')))))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

W(a(a(a(a(a(a(x''''))))))) -> W(a(a(a(a(x'''')))))

The following usable rule for innermost can be oriented:

a(c(d(x))) -> c(x)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = x₁

POL(d(x₁)) = x₁

POL(W(x₁)) = 1 + x₁

POL(a(x₁)) = 1 + x₁

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

W(x₁) -> W(x₁)
a(x₁) -> a(x₁)
c(x₁) -> c(x₁)
d(x₁) -> d(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(c(x₁))	= x₁
POL(V(x₁))	= 1 + x₁
POL(d(x₁))	= x₁
POL(a(x₁))	= 1 + x₁