Termination Proof using AProVE

Term Rewriting System R:
[X, X1, X2]
a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_F}(s(0)) -> A_{_F}(a_{_p}(s(0)))
A_{_F}(s(0)) -> A_{_P}(s(0))
MARK(f(X)) -> A_{_F}(mark(X))
MARK(f(X)) -> MARK(X)
MARK(p(X)) -> A_{_P}(mark(X))
MARK(p(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Polo

Dependency Pair:

A_{_F}(s(0)) -> A_{_F}(a_{_p}(s(0)))

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

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

A_{_F}(s(0)) -> A_{_F}(a_{_p}(s(0)))

two new Dependency Pairs are created:

A_{_F}(s(0)) -> A_{_F}(0)
A_{_F}(s(0)) -> A_{_F}(p(s(0)))

The transformation is resulting in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)
MARK(f(X)) -> MARK(X)

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(f(X)) -> MARK(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = x₁

POL(f(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 3

             ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(s(X)) -> MARK(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pair:

MARK(p(X)) -> MARK(X)

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(p(X)) -> MARK(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

Rules:

a_{_f}(0) -> cons(0, f(s(0)))
a_{_f}(s(0)) -> a_{_f}(a_{_p}(s(0)))
a_{_f}(X) -> f(X)
a_{_p}(s(0)) -> 0
a_{_p}(X) -> p(X)
mark(f(X)) -> a_{_f}(mark(X))
mark(p(X)) -> a_{_p}(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(MARK(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁
POL(s(x₁))	= x₁
POL(f(x₁))	= 1 + x₁
POL(p(x₁))	= x₁