Term Rewriting System R:
[X, Y, X1, X2]
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFROM(X) -> MARK(X)
ALENGTH(cons(X, Y)) -> ALENGTH1(Y)
ALENGTH1(X) -> ALENGTH(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(length(X)) -> ALENGTH(X)
MARK(length1(X)) -> ALENGTH1(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳Neg POLO


Dependency Pairs:

ALENGTH1(X) -> ALENGTH(X)
ALENGTH(cons(X, Y)) -> ALENGTH1(Y)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0


Strategy:

innermost




As we are in the innermost case, we can delete all 14 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳Neg POLO


Dependency Pairs:

ALENGTH1(X) -> ALENGTH(X)
ALENGTH(cons(X, Y)) -> ALENGTH1(Y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. ALENGTH1(X) -> ALENGTH(X)
  2. ALENGTH(cons(X, Y)) -> ALENGTH1(Y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
{2} , {2}
1>1

which lead(s) to this/these maximal multigraph(s):
{2} , {1}
1>1
{1} , {2}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Negative Polynomial Order


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AFROM(X) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0


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)) -> AFROM(mark(X))


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

mark(s(X)) -> s(mark(X))
mark(length(X)) -> alength(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
afrom(X) -> cons(mark(X), from(s(X)))
alength(X) -> length(X)
afrom(X) -> from(X)
alength(nil) -> 0
alength1(X) -> alength(X)
alength(cons(X, Y)) -> s(alength1(Y))
mark(0) -> 0
mark(length1(X)) -> alength1(X)


Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x1) ) = x1

POL( cons(x1, x2) ) = x1 + 1

POL( from(x1) ) = x1 + 1

POL( AFROM(x1) ) = x1

POL( s(x1) ) = x1

POL( mark(x1) ) = x1

POL( length(x1) ) = 0

POL( alength(x1) ) = 0

POL( nil ) = 0

POL( alength1(x1) ) = 0

POL( length1(x1) ) = 0

POL( afrom(x1) ) = x1 + 1

POL( 0 ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Neg POLO
           →DP Problem 4
             ↳Dependency Graph


Dependency Pairs:

MARK(s(X)) -> MARK(X)
AFROM(X) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Neg POLO
           →DP Problem 4
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Usable Rules (Innermost)


Dependency Pair:

MARK(s(X)) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0


Strategy:

innermost




As we are in the innermost case, we can delete all 14 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Neg POLO
           →DP Problem 4
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Size-Change Principle


Dependency Pair:

MARK(s(X)) -> MARK(X)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MARK(s(X)) -> MARK(X)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

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