that the cappuccino is a Omdurman and it is a kind of a pennant is not true.
sent1: if the neutrino does not conventionalize Richardson then the cappuccino is not a pennant and it is not a Omdurman. sent2: something conventionalizes Richardson if it is a doghouse. sent3: something is a Omdurman if it fears hologram. sent4: the cappuccino is a reference if the granny is not a doghouse. sent5: something is a pennant if it conventionalizes Richardson. sent6: the churchwarden is a pennant. sent7: the cappuccino conventionalizes Richardson and it is a doghouse. sent8: the granny is not a reference. sent9: if the granny does not reference then the cappuccino fears hologram.
sent9 & sent8 -> int1: the cappuccino fears hologram.; sent3 -> int2: if the cappuccino does fear hologram it is a Omdurman.; int1 & int2 -> int3: the cappuccino is a Omdurman.; sent5 -> int4: if the cappuccino conventionalizes Richardson then the fact that it is a pennant is not false.; sent7 -> int5: that the cappuccino conventionalizes Richardson is not incorrect.; int4 & int5 -> int6: the cappuccino is a pennant.; int3 & int6 -> hypothesis;
the lemonade does fear hologram.
$hypothesis$ = that the cappuccino is a Omdurman and it is a kind of a pennant is not true. ; $context$ = sent1: if the neutrino does not conventionalize Richardson then the cappuccino is not a pennant and it is not a Omdurman. sent2: something conventionalizes Richardson if it is a doghouse. sent3: something is a Omdurman if it fears hologram. sent4: the cappuccino is a reference if the granny is not a doghouse. sent5: something is a pennant if it conventionalizes Richardson. sent6: the churchwarden is a pennant. sent7: the cappuccino conventionalizes Richardson and it is a doghouse. sent8: the granny is not a reference. sent9: if the granny does not reference then the cappuccino fears hologram. ; $proof$ = 
sent9 & sent8 -> int1: the cappuccino fears hologram.; sent3 -> int2: if the cappuccino does fear hologram it is a Omdurman.; int1 & int2 -> int3: the cappuccino is a Omdurman.; sent5 -> int4: if the cappuccino conventionalizes Richardson then the fact that it is a pennant is not false.; sent7 -> int5: that the cappuccino conventionalizes Richardson is not incorrect.; int4 & int5 -> int6: the cappuccino is a pennant.; int3 & int6 -> hypothesis; __DISPROVED__