that if the fact that the plate is a chartist is true that the openwork does prime birdie is not wrong is wrong.
sent1: there exists nothing that is a gang but not non-multilingual. sent2: the plate is a kind of a chartist if it is a kind of a yeast. sent3: the fact that the plate is a swarm but not a prickliness is not correct. sent4: if the fact that the openwork does not prime birdie is correct that the plate is not a swarm and it is not a kind of a prickliness is not right. sent5: something is a kind of a yeast if that it is not a kind of a swarm and it is not a prickliness is not correct. sent6: a swarm is a yeast. sent7: if something is a yeast then it is not a kind of a chartist. sent8: if the openwork does not prime birdie then that the plate is a swarm but it is not a prickliness is false. sent9: if something is not a yeast and it does not prime birdie it is not a kind of a swarm. sent10: if something is a yeast it is a chartist. sent11: if the plate is a kind of a swarm then it is a yeast.
void -> assump1: Let's assume that the openwork does not prime birdie.; sent4 & assump1 -> int1: that the plate is not a swarm and it is not a prickliness is not true.; sent5 -> int2: the plate is a yeast if the fact that it is not a swarm and it is not a kind of a prickliness is not true.; int1 & int2 -> int3: the plate is a yeast.; sent7 -> int4: the plate is not a chartist if it is a yeast.; int3 & int4 -> int5: the plate is not a kind of a chartist.; void -> assump2: Let's assume that the plate is a chartist.; int5 & assump2 -> int6: this is contradiction.; [assump1] & int6 -> int7: the openwork does prime birdie.; [assump2] & int7 -> hypothesis;
the racketeer is not a kind of a swarm.
sent9 -> int8: the racketeer is not a kind of a swarm if it is not a yeast and does not prime birdie.; sent1 -> int9: the fact that the plate is a gang and it is multilingual is not true.; int9 -> int10: there exists something such that the fact that it gangs and is multilingual does not hold.;
$hypothesis$ = that if the fact that the plate is a chartist is true that the openwork does prime birdie is not wrong is wrong. ; $context$ = sent1: there exists nothing that is a gang but not non-multilingual. sent2: the plate is a kind of a chartist if it is a kind of a yeast. sent3: the fact that the plate is a swarm but not a prickliness is not correct. sent4: if the fact that the openwork does not prime birdie is correct that the plate is not a swarm and it is not a kind of a prickliness is not right. sent5: something is a kind of a yeast if that it is not a kind of a swarm and it is not a prickliness is not correct. sent6: a swarm is a yeast. sent7: if something is a yeast then it is not a kind of a chartist. sent8: if the openwork does not prime birdie then that the plate is a swarm but it is not a prickliness is false. sent9: if something is not a yeast and it does not prime birdie it is not a kind of a swarm. sent10: if something is a yeast it is a chartist. sent11: if the plate is a kind of a swarm then it is a yeast. ; $proof$ = 
void -> assump1: Let's assume that the openwork does not prime birdie.; sent4 & assump1 -> int1: that the plate is not a swarm and it is not a prickliness is not true.; sent5 -> int2: the plate is a yeast if the fact that it is not a swarm and it is not a kind of a prickliness is not true.; int1 & int2 -> int3: the plate is a yeast.; sent7 -> int4: the plate is not a chartist if it is a yeast.; int3 & int4 -> int5: the plate is not a kind of a chartist.; void -> assump2: Let's assume that the plate is a chartist.; int5 & assump2 -> int6: this is contradiction.; [assump1] & int6 -> int7: the openwork does prime birdie.; [assump2] & int7 -> hypothesis; __DISPROVED__