Term Rewriting System R:
[X, Y]
nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

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, Y)) -> INCR(Y)


Rules:


nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y


Strategy:

innermost




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

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

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

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


Rules:


nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y


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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:

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


Rules:


nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x2  
  POL(INCR(x1))=  1 + x1  

resulting in one new DP problem. 



   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:


nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y


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