Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, X3, Z]
a_{_and}(true, X) -> mark(X)
a_{_and}(false, Y) -> false
a_{_and}(X1, X2) -> and(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(and(X1, X2)) -> a_{_and}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(add(X1, X2)) -> a_{_add}(mark(X1), X2)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_AND}(true, X) -> MARK(X)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_IF}(false, X, Y) -> MARK(Y)
A_{_ADD}(0, X) -> MARK(X)
MARK(and(X1, X2)) -> A_{_AND}(mark(X1), X2)
MARK(and(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), X2)
MARK(add(X1, X2)) -> MARK(X1)
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}(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

A_{_IF}(false, X, Y) -> MARK(Y)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(true, X, Y) -> MARK(X)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(and(X1, X2)) -> MARK(X1)
MARK(and(X1, X2)) -> A_{_AND}(mark(X1), X2)
A_{_AND}(true, X) -> MARK(X)

Rules:

a_{_and}(true, X) -> mark(X)
a_{_and}(false, Y) -> false
a_{_and}(X1, X2) -> and(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(and(X1, X2)) -> a_{_and}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(add(X1, X2)) -> a_{_add}(mark(X1), X2)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

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

MARK(and(X1, X2)) -> A_{_AND}(mark(X1), X2)

11 new Dependency Pairs are created:

MARK(and(and(X1'', X2''), X2)) -> A_{_AND}(a_{_and}(mark(X1''), X2''), X2)
MARK(and(if(X1'', X2'', X3'), X2)) -> A_{_AND}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(and(add(X1'', X2''), X2)) -> A_{_AND}(a_{_add}(mark(X1''), X2''), X2)
MARK(and(first(X1'', X2''), X2)) -> A_{_AND}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(and(from(X'), X2)) -> A_{_AND}(a_{_from}(X'), X2)
MARK(and(true, X2)) -> A_{_AND}(true, X2)
MARK(and(false, X2)) -> A_{_AND}(false, X2)
MARK(and(0, X2)) -> A_{_AND}(0, X2)
MARK(and(s(X'), X2)) -> A_{_AND}(s(X'), X2)
MARK(and(nil, X2)) -> A_{_AND}(nil, X2)
MARK(and(cons(X1'', X2''), X2)) -> A_{_AND}(cons(X1'', X2''), X2)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(and(true, X2)) -> A_{_AND}(true, X2)
MARK(and(from(X'), X2)) -> A_{_AND}(a_{_from}(X'), X2)
MARK(and(first(X1'', X2''), X2)) -> A_{_AND}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(and(add(X1'', X2''), X2)) -> A_{_AND}(a_{_add}(mark(X1''), X2''), X2)
MARK(and(if(X1'', X2'', X3'), X2)) -> A_{_AND}(a_{_if}(mark(X1''), X2'', X3'), X2)
A_{_AND}(true, X) -> MARK(X)
MARK(and(and(X1'', X2''), X2)) -> A_{_AND}(a_{_and}(mark(X1''), X2''), X2)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(true, X, Y) -> MARK(X)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(and(X1, X2)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)

Rules:

a_{_and}(true, X) -> mark(X)
a_{_and}(false, Y) -> false
a_{_and}(X1, X2) -> and(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(and(X1, X2)) -> a_{_and}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(add(X1, X2)) -> a_{_add}(mark(X1), X2)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

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

MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)

11 new Dependency Pairs are created:

MARK(if(and(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_and}(mark(X1''), X2''), X2, X3)
MARK(if(if(X1'', X2'', X3''), X2, X3)) -> A_{_IF}(a_{_if}(mark(X1''), X2'', X3''), X2, X3)
MARK(if(add(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_add}(mark(X1''), X2''), X2, X3)
MARK(if(first(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_first}(mark(X1''), mark(X2'')), X2, X3)
MARK(if(from(X'), X2, X3)) -> A_{_IF}(a_{_from}(X'), X2, X3)
MARK(if(true, X2, X3)) -> A_{_IF}(true, X2, X3)
MARK(if(false, X2, X3)) -> A_{_IF}(false, X2, X3)
MARK(if(0, X2, X3)) -> A_{_IF}(0, X2, X3)
MARK(if(s(X'), X2, X3)) -> A_{_IF}(s(X'), X2, X3)
MARK(if(nil, X2, X3)) -> A_{_IF}(nil, X2, X3)
MARK(if(cons(X1'', X2''), X2, X3)) -> A_{_IF}(cons(X1'', X2''), X2, X3)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(if(false, X2, X3)) -> A_{_IF}(false, X2, X3)
MARK(if(true, X2, X3)) -> A_{_IF}(true, X2, X3)
MARK(if(from(X'), X2, X3)) -> A_{_IF}(a_{_from}(X'), X2, X3)
MARK(if(first(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_first}(mark(X1''), mark(X2'')), X2, X3)
MARK(if(add(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_add}(mark(X1''), X2''), X2, X3)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(if(X1'', X2'', X3''), X2, X3)) -> A_{_IF}(a_{_if}(mark(X1''), X2'', X3''), X2, X3)
A_{_IF}(true, X, Y) -> MARK(X)
MARK(if(and(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_and}(mark(X1''), X2''), X2, X3)
MARK(and(from(X'), X2)) -> A_{_AND}(a_{_from}(X'), X2)
MARK(and(first(X1'', X2''), X2)) -> A_{_AND}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(and(add(X1'', X2''), X2)) -> A_{_AND}(a_{_add}(mark(X1''), X2''), X2)
MARK(and(if(X1'', X2'', X3'), X2)) -> A_{_AND}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(and(and(X1'', X2''), X2)) -> A_{_AND}(a_{_and}(mark(X1''), X2''), X2)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(and(X1, X2)) -> MARK(X1)
A_{_AND}(true, X) -> MARK(X)
MARK(and(true, X2)) -> A_{_AND}(true, X2)

Rules:

a_{_and}(true, X) -> mark(X)
a_{_and}(false, Y) -> false
a_{_and}(X1, X2) -> and(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(and(X1, X2)) -> a_{_and}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(add(X1, X2)) -> a_{_add}(mark(X1), X2)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

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

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

11 new Dependency Pairs are created:

MARK(add(and(X1'', X2''), X2)) -> A_{_ADD}(a_{_and}(mark(X1''), X2''), X2)
MARK(add(if(X1'', X2'', X3'), X2)) -> A_{_ADD}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(add(add(X1'', X2''), X2)) -> A_{_ADD}(a_{_add}(mark(X1''), X2''), X2)
MARK(add(first(X1'', X2''), X2)) -> A_{_ADD}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(add(from(X'), X2)) -> A_{_ADD}(a_{_from}(X'), X2)
MARK(add(true, X2)) -> A_{_ADD}(true, X2)
MARK(add(false, X2)) -> A_{_ADD}(false, X2)
MARK(add(0, X2)) -> A_{_ADD}(0, X2)
MARK(add(s(X'), X2)) -> A_{_ADD}(s(X'), X2)
MARK(add(nil, X2)) -> A_{_ADD}(nil, X2)
MARK(add(cons(X1'', X2''), X2)) -> A_{_ADD}(cons(X1'', X2''), X2)

The transformation is resulting 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(add(0, X2)) -> A_{_ADD}(0, X2)
MARK(add(from(X'), X2)) -> A_{_ADD}(a_{_from}(X'), X2)
MARK(add(first(X1'', X2''), X2)) -> A_{_ADD}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(add(add(X1'', X2''), X2)) -> A_{_ADD}(a_{_add}(mark(X1''), X2''), X2)
MARK(add(if(X1'', X2'', X3'), X2)) -> A_{_ADD}(a_{_if}(mark(X1''), X2'', X3'), X2)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(and(X1'', X2''), X2)) -> A_{_ADD}(a_{_and}(mark(X1''), X2''), X2)
MARK(if(true, X2, X3)) -> A_{_IF}(true, X2, X3)
MARK(if(from(X'), X2, X3)) -> A_{_IF}(a_{_from}(X'), X2, X3)
MARK(if(first(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_first}(mark(X1''), mark(X2'')), X2, X3)
MARK(if(add(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_add}(mark(X1''), X2''), X2, X3)
MARK(if(if(X1'', X2'', X3''), X2, X3)) -> A_{_IF}(a_{_if}(mark(X1''), X2'', X3''), X2, X3)
A_{_IF}(true, X, Y) -> MARK(X)
MARK(if(and(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_and}(mark(X1''), X2''), X2, X3)
MARK(and(true, X2)) -> A_{_AND}(true, X2)
MARK(and(from(X'), X2)) -> A_{_AND}(a_{_from}(X'), X2)
MARK(and(first(X1'', X2''), X2)) -> A_{_AND}(a_{_first}(mark(X1''), mark(X2'')), X2)
MARK(and(add(X1'', X2''), X2)) -> A_{_AND}(a_{_add}(mark(X1''), X2''), X2)
MARK(and(if(X1'', X2'', X3'), X2)) -> A_{_AND}(a_{_if}(mark(X1''), X2'', X3'), X2)
A_{_AND}(true, X) -> MARK(X)
MARK(and(and(X1'', X2''), X2)) -> A_{_AND}(a_{_and}(mark(X1''), X2''), X2)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(and(X1, X2)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(false, X2, X3)) -> A_{_IF}(false, X2, X3)

Rules:

a_{_and}(true, X) -> mark(X)
a_{_and}(false, Y) -> false
a_{_and}(X1, X2) -> and(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_first}(0, X) -> nil
a_{_first}(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
a_{_first}(X1, X2) -> first(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(and(X1, X2)) -> a_{_and}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(add(X1, X2)) -> a_{_add}(mark(X1), X2)
mark(first(X1, X2)) -> a_{_first}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

Termination of R could not be shown.
Duration:
0:36 minutes