Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FROM}(X) -> MARK(X)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(sel(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
A_{_FROM}(X) -> MARK(X)

Rules:

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))

10 new Dependency Pairs are created:

A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(mark(X'')), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(cons(X1', X2')), cons(Y, Z)) -> A_{_SEL}(cons(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(mark(X'')), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(mark(X'')))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, s(X''))) -> A_{_SEL}(mark(X), s(mark(X'')))
A_{_SEL}(s(X), cons(Y, 0)) -> A_{_SEL}(mark(X), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(mark(X'')))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(mark(X'')), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(mark(X'')), mark(Z))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)

Rules:

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))

10 new Dependency Pairs are created:

MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(mark(X')), mark(X2))
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
MARK(sel(cons(X1'', X2''), X2)) -> A_{_SEL}(cons(mark(X1''), X2''), mark(X2))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(mark(X')), mark(X2))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(mark(X')))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(mark(X1''), X2''))
MARK(sel(X1, s(X'))) -> A_{_SEL}(mark(X1), s(mark(X')))
MARK(sel(X1, 0)) -> A_{_SEL}(mark(X1), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Negative Polynomial Order

Dependency Pairs:

MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(mark(X1''), X2''))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(mark(X')))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(mark(X'')))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(mark(X'')), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(mark(X'')), mark(Z))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(mark(X')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(mark(X')), mark(X2))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(mark(X1'), X2'))

Rules:

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The following Dependency Pairs can be strictly oriented using the given order.

A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(mark(X'')), mark(Z))
MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(mark(X')), mark(X2))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_sel}(0, cons(X, Y)) -> mark(X)
mark(s(X)) -> s(mark(X))
a_{_from}(X) -> from(X)
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0
mark(from(X)) -> a_{_from}(mark(X))
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))

Used ordering:
Polynomial Order with Interpretation:

POL( A_{_SEL}(x₁, x₂) ) = x₁

POL( s(x₁) ) = 1

POL( a_{_from}(x₁) ) = 0

POL( A_{_FROM}(x₁) ) = 1

POL( MARK(x₁) ) = 1

POL( a_{_sel}(x₁, x₂) ) = 1

POL( mark(x₁) ) = 1

POL( 0 ) = 1

POL( cons(x₁, x₂) ) = 0

POL( from(x₁) ) = 0

POL( sel(x₁, x₂) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(mark(X1''), X2''))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(mark(X')))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(mark(X'')))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(mark(X'')), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(mark(X')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(mark(X1'), X2'))

Rules:

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:03 minutes