R
↳Dependency Pair Analysis
SUM(s(x)) -> SUM(x)
SUM1(s(x)) -> SUM1(x)
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
→DP Problem 2
↳Polo
SUM(s(x)) -> SUM(x)
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))
SUM(s(x)) -> SUM(x)
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))
POL(0) = 0 POL(SUM(x1)) = x1 POL(sum1(x1)) = x1 POL(sum(x1)) = 0 POL(s(x1)) = 1 + x1 POL(+(x1, x2)) = 0
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳Polo
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Polynomial Ordering
SUM1(s(x)) -> SUM1(x)
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))
SUM1(s(x)) -> SUM1(x)
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))
POL(0) = 0 POL(sum1(x1)) = x1 POL(sum(x1)) = 0 POL(s(x1)) = 1 + x1 POL(+(x1, x2)) = 0 POL(SUM1(x1)) = x1
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Polo
→DP Problem 4
↳Dependency Graph
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))