that the scone does neglect financing is not wrong.
sent1: the scone does nest scone and is Britannic if the chine is not Britannic. sent2: if the chine is Britannic then it is a circle. sent3: the scone is Britannic. sent4: something is a nibble that does not neglect adit. sent5: there is something such that the fact that it nibbles and does not neglect adit is wrong. sent6: if the chine is not Britannic the scone is Britannic. sent7: there exists something such that the fact that it is a nibble that does neglect adit is false. sent8: if there is something such that that it is a nibble and it does not neglect adit is not correct then the chine does not circle. sent9: the fact that the chine does not circle is correct if there exists something such that the fact that it does neglect adit is not wrong.
void -> assump1: Let's assume that the chine is Britannic.; sent2 & assump1 -> int1: the chine is a circle.; sent5 & sent8 -> int2: the chine is not a kind of a circle.; int1 & int2 -> int3: this is contradiction.; [assump1] & int3 -> int4: the chine is not Britannic.; int4 & sent1 -> int5: the scone nests scone and it is Britannic.; int5 -> int6: the scone does nest scone.;
$hypothesis$ = that the scone does neglect financing is not wrong. ; $context$ = sent1: the scone does nest scone and is Britannic if the chine is not Britannic. sent2: if the chine is Britannic then it is a circle. sent3: the scone is Britannic. sent4: something is a nibble that does not neglect adit. sent5: there is something such that the fact that it nibbles and does not neglect adit is wrong. sent6: if the chine is not Britannic the scone is Britannic. sent7: there exists something such that the fact that it is a nibble that does neglect adit is false. sent8: if there is something such that that it is a nibble and it does not neglect adit is not correct then the chine does not circle. sent9: the fact that the chine does not circle is correct if there exists something such that the fact that it does neglect adit is not wrong. ; $proof$ = 
void -> assump1: Let's assume that the chine is Britannic.; sent2 & assump1 -> int1: the chine is a circle.; sent5 & sent8 -> int2: the chine is not a kind of a circle.; int1 & int2 -> int3: this is contradiction.; [assump1] & int3 -> int4: the chine is not Britannic.; int4 & sent1 -> int5: the scone nests scone and it is Britannic.; int5 -> int6: the scone does nest scone.; __UNKNOWN__