R
↳Dependency Pair Analysis
2ND(cons(X, X1)) -> 2ND(cons1(X, X1))
FROM(X) -> FROM(s(X))
R
↳DPs
→DP Problem 1
↳Instantiation Transformation
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)))
one new Dependency Pair is created:
FROM(X) -> FROM(s(X))
FROM(s(X'')) -> FROM(s(s(X'')))
R
↳DPs
→DP Problem 1
↳Inst
→DP Problem 2
↳Instantiation Transformation
FROM(s(X'')) -> FROM(s(s(X'')))
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
from(X) -> cons(X, from(s(X)))
one new Dependency Pair is created:
FROM(s(X'')) -> FROM(s(s(X'')))
FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
R
↳DPs
→DP Problem 1
↳Inst
→DP Problem 2
↳Inst
...
→DP Problem 3
↳Instantiation Transformation
FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
from(X) -> cons(X, from(s(X)))
one new Dependency Pair is created:
FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
R
↳DPs
→DP Problem 1
↳Inst
→DP Problem 2
↳Inst
...
→DP Problem 4
↳Instantiation Transformation
FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
from(X) -> cons(X, from(s(X)))
one new Dependency Pair is created:
FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
R
↳DPs
→DP Problem 1
↳Inst
→DP Problem 2
↳Inst
...
→DP Problem 5
↳Instantiation Transformation
FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
from(X) -> cons(X, from(s(X)))
one new Dependency Pair is created:
FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))
R
↳DPs
→DP Problem 1
↳Inst
→DP Problem 2
↳Inst
...
→DP Problem 6
↳Remaining Obligation(s)
FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
from(X) -> cons(X, from(s(X)))