Termination Proof using AProVE

Term Rewriting System R:
[Z, X, Y, X1, X2]
a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
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_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)
A_{_FROM}(X) -> MARK(X)
A_{_ADD}(0, X) -> MARK(X)
MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X2)
MARK(len(X)) -> A_{_LEN}(mark(X))
MARK(len(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(len(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))
A_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
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_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)

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

mark(len(X)) -> a_{_len}(mark(X))
mark(nil) -> nil
mark(s(X)) -> s(X)
a_{_fst}(0, Z) -> nil
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_len}(X) -> len(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(from(X)) -> a_{_from}(mark(X))
a_{_len}(nil) -> 0
a_{_add}(X1, X2) -> add(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
mark(0) -> 0
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_len}(cons(X, Z)) -> s(len(Z))

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( cons(x₁, x₂) ) = x₁ + 1

POL( A_{_FROM}(x₁) ) = x₁

POL( len(x₁) ) = x₁

POL( A_{_ADD}(x₁, x₂) ) = x₂

POL( add(x₁, x₂) ) = x₁ + x₂

POL( mark(x₁) ) = x₁

POL( A_{_FST}(x₁, x₂) ) = x₂

POL( fst(x₁, x₂) ) = x₁ + x₂

POL( from(x₁) ) = x₁ + 1

POL( a_{_len}(x₁) ) = x₁

POL( nil ) = 0

POL( s(x₁) ) = 0

POL( a_{_fst}(x₁, x₂) ) = x₁ + x₂

POL( a_{_from}(x₁) ) = x₁ + 1

POL( 0 ) = 0

POL( a_{_add}(x₁, x₂) ) = x₁ + x₂

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

MARK(len(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
A_{_FROM}(X) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
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

                 ↳Negative Polynomial Order

Dependency Pairs:

MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Strategy:

innermost

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

MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))

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

mark(len(X)) -> a_{_len}(mark(X))
mark(nil) -> nil
mark(s(X)) -> s(X)
a_{_fst}(0, Z) -> nil
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_len}(X) -> len(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(from(X)) -> a_{_from}(mark(X))
a_{_len}(nil) -> 0
a_{_add}(X1, X2) -> add(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
mark(0) -> 0
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_len}(cons(X, Z)) -> s(len(Z))

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₁ + x₂ + 1

POL( A_{_ADD}(x₁, x₂) ) = x₂

POL( mark(x₁) ) = x₁

POL( len(x₁) ) = x₁

POL( fst(x₁, x₂) ) = x₁ + x₂

POL( a_{_len}(x₁) ) = x₁

POL( nil ) = 0

POL( s(x₁) ) = 0

POL( a_{_fst}(x₁, x₂) ) = x₁ + x₂

POL( cons(x₁, x₂) ) = 0

POL( from(x₁) ) = 0

POL( a_{_from}(x₁) ) = 0

POL( 0 ) = 0

POL( a_{_add}(x₁, x₂) ) = x₁ + x₂ + 1

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

A_{_ADD}(0, X) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
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 5

                 ↳Usable Rules (Innermost)

Dependency Pairs:

MARK(len(X)) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Strategy:

innermost

As we are in the innermost case, we can delete all 19 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Size-Change Principle

Dependency Pairs:

MARK(len(X)) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

MARK(len(X)) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)

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:

len(x₁) -> len(x₁)
fst(x₁, x₂) -> fst(x₁, x₂)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:05 minutes