Rami Botros

"Love is when things work out." - Rami Botros

"Well, you can limit the universe using the operator '~=' to compare objects you're working on as an extra condition. I'll give you the solution without extra explanation, but if you don't get the concept, feel free to ask. A x ( Square(x) => ( A y (Triangle(y) => LeftOf(y,x)) \/ ( A y(Triangle(y) => LeftOf(x,y)) /\ A z( (Square(z) /\ ~x=z) => LeftOf(x,z) ) ) ) )" - Rami Botros

"How does the syntax go?" - Rami Botros

"It would be resolved if you can play around with the definitions of the universes you're investigating. You can always say "For all squares in the universe all_Squares" , but can you specify different universes? Here I'm assuming I can use a property of objects called HPos (horizontal position). For each SquareA in all_Squares { Disjunction { For each TriangleA in all_Triangles{ SquareA.HPos > TriangleA.HPos } ; Conjunction{ For each TriangleA in all_Triangles{ SquareA.HPos < TriangleA.HPos } ; For each SquareB in *rest_of_the_Squares*{ SquareA.HPos < SquareB.HPos } } } }" - Rami Botros

"You want to prove that [n-1] breaks result in [n] pieces. In another formulation: [PiecesAfterBreaks(n-1)=n] Base case: For [n=1] it works, because if you don't break the bar at all, you're left with what you originally had - one bar. [PiecesAfterBreaks(0)=1] For the step case [P(n+1)], it works because after one break of a bar you get a bar which resembles the bar you just broke, plus an additional bar. [PiecesAfterBreaks(n+1)=PiecesAfterBreaks(n)+1]. This equation holds down until [n=0]" - Rami Botros

"A scientific pocket calculator." - Rami Botros

"The only reason you were so highly up-modded is that people liked your joke." - Rami Botros

"English, motherfucker! Do you speak it?" - Rami Botros