: for all natural numbers a and b, if a > b then a^2 − b^2 is composite.
is it?
∀a, b,∈N, a > b−→a2−b2 AND composite(ab)?
or would you have to add a 3rd variable c and let c = a^2-b^2? and use composite(c)
Translate this to predicate logic
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predicate-logic
logic-translation
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0composite($a^2 - b^2$) makes more sense. – 2017-02-15
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0You have already formalized it [here](http://math.stackexchange.com/questions/2143678/composite-proof). – 2017-02-15
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0^ did you see my last comment about my confusion... my translation wasn't perfect predicate logic but i think martins translation is more accurate so ^(and) composite(a^2−b^2)? – 2017-02-15