there is something such that if it is not a gill the fact that it does not purify antilogarithm and is a asterisked is false.
sent1: the recognition does not purify antilogarithm and is a kind of a asterisked if it is not a gill. sent2: if the recognition is a asterisked it is a gill. sent3: there exists something such that if it is a gill the fact that the fact that it does not purify antilogarithm and is a asterisked is not incorrect does not hold. sent4: that the recognition does purify antilogarithm and it is a asterisked is incorrect. sent5: the recognition does not purify antilogarithm. sent6: there is something such that if it is not a gill it does not purify antilogarithm and it is a asterisked.
void -> assump2: Let's assume that the recognition does not purify antilogarithm but it is a asterisked.; assump2 -> int1: the recognition is a asterisked.; int1 & sent2 -> int2: the recognition is a gill.; void -> assump1: Let's assume that the recognition is not a gill.; int2 & assump1 -> int3: this is contradiction.; [assump2] & int3 -> int4: that the recognition does not purify antilogarithm and is a asterisked is wrong.; [assump1] & int4 -> int5: the fact that if the recognition is not a kind of the gill the fact that the recognition does not purify antilogarithm and is a kind of a asterisked does not hold is not incorrect.; int5 -> hypothesis;
$hypothesis$ = there is something such that if it is not a gill the fact that it does not purify antilogarithm and is a asterisked is false. ; $context$ = sent1: the recognition does not purify antilogarithm and is a kind of a asterisked if it is not a gill. sent2: if the recognition is a asterisked it is a gill. sent3: there exists something such that if it is a gill the fact that the fact that it does not purify antilogarithm and is a asterisked is not incorrect does not hold. sent4: that the recognition does purify antilogarithm and it is a asterisked is incorrect. sent5: the recognition does not purify antilogarithm. sent6: there is something such that if it is not a gill it does not purify antilogarithm and it is a asterisked. ; $proof$ = 
void -> assump2: Let's assume that the recognition does not purify antilogarithm but it is a asterisked.; assump2 -> int1: the recognition is a asterisked.; int1 & sent2 -> int2: the recognition is a gill.; void -> assump1: Let's assume that the recognition is not a gill.; int2 & assump1 -> int3: this is contradiction.; [assump2] & int3 -> int4: that the recognition does not purify antilogarithm and is a asterisked is wrong.; [assump1] & int4 -> int5: the fact that if the recognition is not a kind of the gill the fact that the recognition does not purify antilogarithm and is a kind of a asterisked does not hold is not incorrect.; int5 -> hypothesis; __PROVED__