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