Term Rewriting System R:
[X, L]
incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

INCR(cons(X, L)) -> INCR(L)
ADX(cons(X, L)) -> INCR(cons(X, adx(L)))
ADX(cons(X, L)) -> ADX(L)
NATS -> ADX(zeros)
NATS -> ZEROS
ZEROS -> ZEROS

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:

INCR(cons(X, L)) -> INCR(L)


Rules:


incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L


Strategy:

innermost




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

INCR(cons(X, L)) -> INCR(L)
one new Dependency Pair is created:

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))


Rules:


incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L


Strategy:

innermost




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

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))
one new Dependency Pair is created:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))


Rules:


incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L


Strategy:

innermost




The following dependency pair can be strictly oriented:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))


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:
INCR(x1) -> INCR(x1)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes