Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FIRST}(s(X), cons(Y, Z)) -> MARK(Y)
A_{_FROM}(X) -> MARK(X)
MARK(first(X1, X2)) -> A_{_FIRST}(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Negative Polynomial Order

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> A_{_FIRST}(mark(X1), mark(X2))
A_{_FIRST}(s(X), cons(Y, Z)) -> MARK(Y)

Rules:

a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Strategy:

innermost

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

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

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

a_{_first}(0, X) -> nil
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
mark(s(X)) -> s(mark(X))
a_{_from}(X) -> from(X)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(nil) -> nil
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
a_{_first}(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
mark(from(X)) -> a_{_from}(mark(X))

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( cons(x₁, x₂) ) = x₁ + 1

POL( from(x₁) ) = x₁ + 1

POL( first(x₁, x₂) ) = x₁ + x₂

POL( A_{_FROM}(x₁) ) = x₁

POL( s(x₁) ) = x₁

POL( A_{_FIRST}(x₁, x₂) ) = x₂

POL( mark(x₁) ) = x₁

POL( a_{_first}(x₁, x₂) ) = x₁ + x₂

POL( 0 ) = 0

POL( nil ) = 0

POL( a_{_from}(x₁) ) = x₁ + 1

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

MARK(s(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> A_{_FIRST}(mark(X1), mark(X2))

Rules:

a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Usable Rules (Innermost)

Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Rules:

a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Strategy:

innermost

As we are in the innermost case, we can delete all 11 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Size-Change Principle

Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

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

{1, 2, 3}	,	{1, 2, 3}
1	>	1

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

{1, 2, 3}	,	{1, 2, 3}
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:

first(x₁, x₂) -> first(x₁, x₂)
s(x₁) -> s(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes