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)

Innermost 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

         ↳Argument Filtering and Ordering

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)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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)

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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)
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)
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)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{and, a_{_and}, A_{_AND}} > MARK
{a_{_from}, from}
{add, a_{_add}, A_{_ADD}} > MARK
{add, a_{_add}, A_{_ADD}} > s
{a_{_first}, first}
{a_{_if}, if} > A_{_IF} > MARK
{0, nil}

resulting in one new DP problem.
Used Argument Filtering System:

A_{_AND}(x₁, x₂) -> A_{_AND}(x₁, x₂)
MARK(x₁) -> MARK(x₁)
first(x₁, x₂) -> first(x₁, x₂)
add(x₁, x₂) -> add(x₁, x₂)
A_{_ADD}(x₁, x₂) -> A_{_ADD}(x₁, x₂)
mark(x₁) -> x₁
and(x₁, x₂) -> and(x₁, x₂)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
A_{_IF}(x₁, x₂, x₃) -> A_{_IF}(x₁, x₂, x₃)
a_{_and}(x₁, x₂) -> a_{_and}(x₁, x₂)
a_{_if}(x₁, x₂, x₃) -> a_{_if}(x₁, x₂, x₃)
a_{_add}(x₁, x₂) -> a_{_add}(x₁, x₂)
a_{_first}(x₁, x₂) -> a_{_first}(x₁, x₂)
from(x₁) -> from(x₁)
a_{_from}(x₁) -> a_{_from}(x₁)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 2

             ↳Dependency Graph

Dependency Pair:

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes