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