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↳Dependency Pair Analysis
+'(x, minus(y)) -> MINUS(+(minus(x), y))
+'(x, minus(y)) -> +'(minus(x), y)
+'(x, minus(y)) -> MINUS(x)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
+'(minus(+(x, 1)), 1) -> MINUS(x)
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↳DPs
→DP Problem 1
↳Polynomial Ordering
+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, minus(y)) -> +'(minus(x), y)
minus(0) -> 0
minus(minus(x)) -> x
+(x, 0) -> x
+(0, y) -> y
+(minus(1), 1) -> 0
+(x, minus(y)) -> minus(+(minus(x), y))
+(x, +(y, z)) -> +(+(x, y), z)
+(minus(+(x, 1)), 1) -> minus(x)
+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
minus(0) -> 0
minus(minus(x)) -> x
+(x, 0) -> x
+(0, y) -> y
+(minus(1), 1) -> 0
+(x, minus(y)) -> minus(+(minus(x), y))
+(x, +(y, z)) -> +(+(x, y), z)
+(minus(+(x, 1)), 1) -> minus(x)
POL(0) = 0 POL(1) = 0 POL(minus(x1)) = x1 POL(+(x1, x2)) = 1 + x1 + x2 POL(+'(x1, x2)) = 1 + x2
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↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Remaining Obligation(s)
+'(x, minus(y)) -> +'(minus(x), y)
minus(0) -> 0
minus(minus(x)) -> x
+(x, 0) -> x
+(0, y) -> y
+(minus(1), 1) -> 0
+(x, minus(y)) -> minus(+(minus(x), y))
+(x, +(y, z)) -> +(+(x, y), z)
+(minus(+(x, 1)), 1) -> minus(x)