if that the greenwood is not appointive is true the fact that the dustcart is both not a fitness and an icing does not hold.
sent1: if the greenwood is not appointive it is not a ammine. sent2: the greenwood is not a ammine. sent3: that the dustcart is not a kind of a fitness but it does ice is wrong if the greenwood is not non-appointive. sent4: the dustcart is not a fitness and does ice if the greenwood is not appointive. sent5: there exists something such that it is not a ammine. sent6: if the greenwood is not appointive then it is not a ammine and is aeronautical. sent7: there exists something such that it is not a fitness and it is a ammine. sent8: if the fricandeau does not turn distributary then the fact that the dustcart is not a fitness but it is an icing is false. sent9: if the fricandeau does not turn distributary then the dustcart is not a fitness and is a kind of an icing. sent10: the fricandeau does not turn distributary if something that is not a ammine is aeronautical. sent11: the fricandeau does not turn distributary if there is something such that it is a kind of a ammine and it is aeronautical. sent12: the fact that the dustcart is a kind of a fitness and it ices is not right if the greenwood is not appointive. sent13: if the fricandeau does not turn distributary the fact that the dustcart is a fitness and it is an icing is not true. sent14: that the dustcart is a fitness and ices is false.
void -> assump1: Let's assume that the greenwood is not appointive.; sent6 & assump1 -> int1: the greenwood is not a ammine and not non-aeronautical.; int1 -> int2: there exists something such that it is not a ammine and it is aeronautical.; int2 & sent10 -> int3: the fricandeau does not turn distributary.; sent8 & int3 -> int4: the fact that the dustcart is not a kind of a fitness and ices is not correct.; [assump1] & int4 -> hypothesis;
$hypothesis$ = if that the greenwood is not appointive is true the fact that the dustcart is both not a fitness and an icing does not hold. ; $context$ = sent1: if the greenwood is not appointive it is not a ammine. sent2: the greenwood is not a ammine. sent3: that the dustcart is not a kind of a fitness but it does ice is wrong if the greenwood is not non-appointive. sent4: the dustcart is not a fitness and does ice if the greenwood is not appointive. sent5: there exists something such that it is not a ammine. sent6: if the greenwood is not appointive then it is not a ammine and is aeronautical. sent7: there exists something such that it is not a fitness and it is a ammine. sent8: if the fricandeau does not turn distributary then the fact that the dustcart is not a fitness but it is an icing is false. sent9: if the fricandeau does not turn distributary then the dustcart is not a fitness and is a kind of an icing. sent10: the fricandeau does not turn distributary if something that is not a ammine is aeronautical. sent11: the fricandeau does not turn distributary if there is something such that it is a kind of a ammine and it is aeronautical. sent12: the fact that the dustcart is a kind of a fitness and it ices is not right if the greenwood is not appointive. sent13: if the fricandeau does not turn distributary the fact that the dustcart is a fitness and it is an icing is not true. sent14: that the dustcart is a fitness and ices is false. ; $proof$ = 
void -> assump1: Let's assume that the greenwood is not appointive.; sent6 & assump1 -> int1: the greenwood is not a ammine and not non-aeronautical.; int1 -> int2: there exists something such that it is not a ammine and it is aeronautical.; int2 & sent10 -> int3: the fricandeau does not turn distributary.; sent8 & int3 -> int4: the fact that the dustcart is not a kind of a fitness and ices is not correct.; [assump1] & int4 -> hypothesis; __PROVED__