2nd(cons1(

2nd(cons(

from(

R

↳Dependency Pair Analysis

2ND(cons(X,X1)) -> 2ND(cons1(X,X1))

FROM(X) -> FROM(s(X))

Furthermore,

R

↳DPs

→DP Problem 1

↳Usable Rules (Innermost)

**FROM( X) -> FROM(s(X))**

2nd(cons1(X, cons(Y,Z))) ->Y

2nd(cons(X,X1)) -> 2nd(cons1(X,X1))

from(X) -> cons(X, from(s(X)))

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

R

↳DPs

→DP Problem 1

↳UsableRules

→DP Problem 2

↳Non Termination

**FROM( X) -> FROM(s(X))**

none

innermost

Found an infinite P-chain over R:

P =

FROM(X) -> FROM(s(X))

R = none

s = FROM(

evaluates to t =FROM(s(

Thus, s starts an infinite chain as s matches t.

Duration:

0:01 minutes