R
↳Dependency Pair Analysis
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
FROM(X) -> FROM(s(X))
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
→DP Problem 2
↳Inst
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
innermost
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
POL(cons(x1, x2)) = 0 POL(FIRST(x1, x2)) = x1 POL(s(x1)) = 1 + x1
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳Inst
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
innermost
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Instantiation Transformation
FROM(X) -> FROM(s(X))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
innermost
one new Dependency Pair is created:
FROM(X) -> FROM(s(X))
FROM(s(X'')) -> FROM(s(s(X'')))
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Inst
→DP Problem 4
↳Remaining Obligation(s)
FROM(s(X'')) -> FROM(s(s(X'')))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
innermost