the freestyle occurs and the manhandling workaholic does not occur.
sent1: the zoologicalness does not occur. sent2: the manhandling workaholic does not occur if the fact that that the hypothermicness or the iniquity or both occurs is not false is wrong. sent3: the uninjectableness happens. sent4: if both the manhandling occupation and the uninjectableness occurs then the shearing does not occur. sent5: the nontechnicalness occurs and the shearing occurs. sent6: both the freestyle and the mishap happens if the zoologicalness does not occur. sent7: if the questionableness does not occur that either the middlingness or the kid or both happens is not right.
void -> assump1: Let's assume that the fact that the manhandling occupation occurs is true.; assump1 & sent3 -> int1: both the manhandling occupation and the uninjectableness happens.; int1 & sent4 -> int2: the shearing does not occur.; sent5 -> int3: the shearing happens.; int2 & int3 -> int4: this is contradiction.; [assump1] & int4 -> int5: the manhandling occupation does not occur.; sent6 & sent1 -> int6: the freestyle and the mishap occurs.; int6 -> int7: the freestyle happens.;
the fact that the freestyle occurs but the manhandling workaholic does not occur is not correct.
 -> hypothesis;
$hypothesis$ = the freestyle occurs and the manhandling workaholic does not occur. ; $context$ = sent1: the zoologicalness does not occur. sent2: the manhandling workaholic does not occur if the fact that that the hypothermicness or the iniquity or both occurs is not false is wrong. sent3: the uninjectableness happens. sent4: if both the manhandling occupation and the uninjectableness occurs then the shearing does not occur. sent5: the nontechnicalness occurs and the shearing occurs. sent6: both the freestyle and the mishap happens if the zoologicalness does not occur. sent7: if the questionableness does not occur that either the middlingness or the kid or both happens is not right. ; $proof$ = 
void -> assump1: Let's assume that the fact that the manhandling occupation occurs is true.; assump1 & sent3 -> int1: both the manhandling occupation and the uninjectableness happens.; int1 & sent4 -> int2: the shearing does not occur.; sent5 -> int3: the shearing happens.; int2 & int3 -> int4: this is contradiction.; [assump1] & int4 -> int5: the manhandling occupation does not occur.; sent6 & sent1 -> int6: the freestyle and the mishap occurs.; int6 -> int7: the freestyle happens.; __UNKNOWN__