cons(

inf(

R

↳Dependency Pair Analysis

INF(x) -> CONS(x, inf(s(x)))

INF(x) -> INF(s(x))

Furthermore,

R

↳DPs

→DP Problem 1

↳Usable Rules (Innermost)

**INF( x) -> INF(s(x))**

cons(x, cons(y,z)) -> big

inf(x) -> cons(x, inf(s(x)))

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

R

↳DPs

→DP Problem 1

↳UsableRules

→DP Problem 2

↳Non Termination

**INF( x) -> INF(s(x))**

none

innermost

Found an infinite P-chain over R:

P =

INF(x) -> INF(s(x))

R = none

s = INF(

evaluates to t =INF(s(

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

Duration:

0:01 minutes