Term Rewriting System R:
[X, Y, Z, X1, X2]
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(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.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
AFROM(X) -> MARK(X)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(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:

MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)


The following usable rules for innermost can be oriented:

mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{AFIRST, afirst, first} > MARK
{afrom, from} > AFROM > MARK
0 > nil

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
from(x1) -> from(x1)
first(x1, x2) -> first(x1, x2)
AFROM(x1) -> AFROM(x1)
s(x1) -> s(x1)
cons(x1, x2) -> x1
AFIRST(x1, x2) -> AFIRST(x1, x2)
mark(x1) -> x1
afirst(x1, x2) -> afirst(x1, x2)
afrom(x1) -> afrom(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 2
Argument Filtering and Ordering


Dependency Pair:

MARK(cons(X1, X2)) -> MARK(X1)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(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 pair can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
cons(x1, x2) -> cons(x1, x2)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 2
AFS
             ...
               →DP Problem 3
Dependency Graph


Dependency Pair:


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(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 resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:42 minutes