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