Documentation

Bdd.Count

def Count.Solution {n m : } (O : OBdd n m) :
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    @[reducible, inline]
    abbrev Count.numSolutions {n m : } (O : OBdd n m) :
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      def Count.count {n m : } (O : OBdd n m) :
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        theorem Count.count_corrent {n m : } {O : OBdd n m} :