R
↳Dependency Pair Analysis
I(+(x, y)) -> +'(i(x), i(y))
I(+(x, y)) -> I(x)
I(+(x, y)) -> I(y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
R
↳DPs
→DP Problem 1
↳Argument Filtering and Ordering
→DP Problem 2
↳AFS
+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
+'(x, +(y, z)) -> +'(x, y)
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
POL(0) = 0 POL(i(x1)) = x1 POL(+(x1, x2)) = 1 + x1 + x2 POL(+'(x1, x2)) = 1 + x1 + x2
+'(x1, x2) -> +'(x1, x2)
+(x1, x2) -> +(x1, x2)
i(x1) -> i(x1)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 3
↳Argument Filtering and Ordering
→DP Problem 2
↳AFS
+'(x, +(y, z)) -> +'(+(x, y), z)
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
+'(x, +(y, z)) -> +'(+(x, y), z)
POL(+(x1, x2)) = 1 + x1 + x2
+'(x1, x2) -> x2
+(x1, x2) -> +(x1, x2)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 3
↳AFS
...
→DP Problem 4
↳Dependency Graph
→DP Problem 2
↳AFS
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳Argument Filtering and Ordering
I(+(x, y)) -> I(y)
I(+(x, y)) -> I(x)
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
I(+(x, y)) -> I(y)
I(+(x, y)) -> I(x)
POL(I(x1)) = x1 POL(+(x1, x2)) = 1 + x1 + x2
I(x1) -> I(x1)
+(x1, x2) -> +(x1, x2)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 5
↳Dependency Graph
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x