Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z))
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z)))
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
fcons(X, Z) -> cons(X, Z)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Removing Redundant Rules for Innermost Termination

Removing the following rules from R which left hand sides contain non normal subterms

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z))
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z)))

     ↳RRRI

       →TRS2

         ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, n_{_from}(s(X)))
FROM(X) -> S(X)
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
QUOTE(n_{_s}(X)) -> ACTIVATE(X)
QUOTE(n_{_sel}(X, Z)) -> ACTIVATE(X)
QUOTE(n_{_sel}(X, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(X1, X2)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_s}(X)) -> S(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
QUOTE1(n_{_cons}(X, Z)) -> QUOTE(activate(X))
QUOTE1(n_{_cons}(X, Z)) -> ACTIVATE(X)
QUOTE1(n_{_cons}(X, Z)) -> QUOTE1(activate(Z))
QUOTE1(n_{_cons}(X, Z)) -> ACTIVATE(Z)
QUOTE1(n_{_first}(X, Z)) -> ACTIVATE(X)
QUOTE1(n_{_first}(X, Z)) -> ACTIVATE(Z)
UNQUOTE(01) -> 0'
UNQUOTE(s1(X)) -> S(unquote(X))
UNQUOTE(s1(X)) -> UNQUOTE(X)
UNQUOTE1(nil1) -> NIL
UNQUOTE1(cons1(X, Z)) -> FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) -> UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)
FCONS(X, Z) -> CONS(X, Z)

Furthermore, R contains four SCCs.

     ↳RRRI

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Negative Polynomial Order

           →DP Problem 2

             ↳SCP

           →DP Problem 3

             ↳NonTerm

Dependency Pair:

QUOTE(n_{_s}(X)) -> QUOTE(activate(X))

Rules:

sel(X1, X2) -> n_{_sel}(X1, X2)
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
0 -> n_₀
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
nil -> n_{_nil}
fcons(X, Z) -> cons(X, Z)

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

QUOTE(n_{_s}(X)) -> QUOTE(activate(X))

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

activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
sel(X1, X2) -> n_{_sel}(X1, X2)

Used ordering:
Polynomial Order with Interpretation:

POL( QUOTE(x₁) ) = x₁

POL( n_{_s}(x₁) ) = x₁ + 1

POL( activate(x₁) ) = x₁

POL( n_{_first}(x₁, x₂) ) = 0

POL( first(x₁, x₂) ) = 0

POL( n_{_from}(x₁) ) = 0

POL( from(x₁) ) = 0

POL( n_₀ ) = 0

POL( 0 ) = 0

POL( n_{_cons}(x₁, x₂) ) = 0

POL( cons(x₁, x₂) ) = 0

POL( n_{_nil} ) = 0

POL( nil ) = 0

POL( s(x₁) ) = x₁ + 1

POL( n_{_sel}(x₁, x₂) ) = 0

POL( sel(x₁, x₂) ) = 0

This results in one new DP problem.

     ↳RRRI

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Neg POLO

...

               →DP Problem 5

                 ↳Dependency Graph

           →DP Problem 2

             ↳SCP

           →DP Problem 3

             ↳NonTerm

Dependency Pair:

Rules:

sel(X1, X2) -> n_{_sel}(X1, X2)
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
0 -> n_₀
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
nil -> n_{_nil}
fcons(X, Z) -> cons(X, Z)

Using the Dependency Graph resulted in no new DP problems.

     ↳RRRI

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Neg POLO

           →DP Problem 2

             ↳Size-Change Principle

           →DP Problem 3

             ↳NonTerm

Dependency Pair:

UNQUOTE(s1(X)) -> UNQUOTE(X)

Rules:

sel(X1, X2) -> n_{_sel}(X1, X2)
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
0 -> n_₀
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
nil -> n_{_nil}
fcons(X, Z) -> cons(X, Z)

We number the DPs as follows:

UNQUOTE(s1(X)) -> UNQUOTE(X)

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s1(x₁) -> s1(x₁)

We obtain no new DP problems.

     ↳RRRI

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Neg POLO

           →DP Problem 2

             ↳SCP

           →DP Problem 3

             ↳Non Termination

Dependency Pair:

QUOTE1(n_{_cons}(X, Z)) -> QUOTE1(activate(Z))

Rules:

sel(X1, X2) -> n_{_sel}(X1, X2)
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
0 -> n_₀
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
nil -> n_{_nil}
fcons(X, Z) -> cons(X, Z)

Found an infinite P-chain over R:
P =

QUOTE1(n_{_cons}(X, Z)) -> QUOTE1(activate(Z))

R =

sel(X1, X2) -> n_{_sel}(X1, X2)
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
quote(n_₀) -> 01
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_₀) -> 0
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(n_{_nil}) -> nil
activate(n_{_s}(X)) -> s(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(X) -> X
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
quote1(n_{_first}(X, Z)) -> first1(activate(X), activate(Z))
unquote(01) -> 0
unquote(s1(X)) -> s(unquote(X))
0 -> n_₀
unquote1(nil1) -> nil
unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
nil -> n_{_nil}
fcons(X, Z) -> cons(X, Z)

s = QUOTE1(activate(n_{_from}(X''')))
evaluates to t =QUOTE1(activate(n_{_from}(s(X'''))))

Thus, s starts an infinite chain as s matches t.

Innermost Non-Termination of R could be shown.
Duration:
0:06 minutes