the harpulla is not incongruous.
sent1: if the mycophagist is a Polynesian and is not a gross the harpulla is not incongruous. sent2: the mycophagist does not fry Sinatra if that it is invitational and/or a practice hold. sent3: if that something is not circulatory hold then it stages HA and it is a practice. sent4: the harpulla is incongruous if the mycophagist is not incongruous. sent5: something is both Polynesian and not a gross if it is not a skim. sent6: if that that something does not fry Sinatra but it is invitational is not false is not right then it is not invitational. sent7: if the mycophagist is not a practice it is not a skim. sent8: if there is something such that it is a practice that the mycophagist does not fry Sinatra and is invitational is incorrect. sent9: that the HA stages HA and it is circulatory is not true. sent10: if that the plainclothesman either does not fry Sinatra or does not stage HA or both is wrong then the HA is not circulatory. sent11: if something is not invitational it is not shallow. sent12: the fact that everything is circulatory and is a domatium hold. sent13: if the mycophagist does not fry Sinatra that that the plainclothesman does not fry Sinatra and/or does not stage HA is correct does not hold.
void -> assump1: Let's assume that the mycophagist is a practice.; assump1 -> int1: that the mycophagist is invitational and/or it is a practice is right.; int1 & sent2 -> int2: the mycophagist does not fry Sinatra.; int2 & sent13 -> int3: that the plainclothesman does not fry Sinatra or does not stage HA or both is not correct.; int3 & sent10 -> int4: the HA is not circulatory.; sent12 -> int5: the HA is circulatory and it is a kind of a domatium.; int5 -> int6: the HA is circulatory.; int4 & int6 -> int7: this is contradiction.; [assump1] & int7 -> int8: the mycophagist is not a practice.; int8 & sent7 -> int9: the mycophagist is not a skim.; sent5 -> int10: if the mycophagist is not a kind of a skim it is a kind of a Polynesian and it does not gross.; int9 & int10 -> int11: the mycophagist is both a Polynesian and not a gross.; int11 & sent1 -> hypothesis;
the harpulla is incongruous.
sent9 -> int12: there exists something such that the fact that it does stage HA and is circulatory does not hold.;
$hypothesis$ = the harpulla is not incongruous. ; $context$ = sent1: if the mycophagist is a Polynesian and is not a gross the harpulla is not incongruous. sent2: the mycophagist does not fry Sinatra if that it is invitational and/or a practice hold. sent3: if that something is not circulatory hold then it stages HA and it is a practice. sent4: the harpulla is incongruous if the mycophagist is not incongruous. sent5: something is both Polynesian and not a gross if it is not a skim. sent6: if that that something does not fry Sinatra but it is invitational is not false is not right then it is not invitational. sent7: if the mycophagist is not a practice it is not a skim. sent8: if there is something such that it is a practice that the mycophagist does not fry Sinatra and is invitational is incorrect. sent9: that the HA stages HA and it is circulatory is not true. sent10: if that the plainclothesman either does not fry Sinatra or does not stage HA or both is wrong then the HA is not circulatory. sent11: if something is not invitational it is not shallow. sent12: the fact that everything is circulatory and is a domatium hold. sent13: if the mycophagist does not fry Sinatra that that the plainclothesman does not fry Sinatra and/or does not stage HA is correct does not hold. ; $proof$ = 
void -> assump1: Let's assume that the mycophagist is a practice.; assump1 -> int1: that the mycophagist is invitational and/or it is a practice is right.; int1 & sent2 -> int2: the mycophagist does not fry Sinatra.; int2 & sent13 -> int3: that the plainclothesman does not fry Sinatra or does not stage HA or both is not correct.; int3 & sent10 -> int4: the HA is not circulatory.; sent12 -> int5: the HA is circulatory and it is a kind of a domatium.; int5 -> int6: the HA is circulatory.; int4 & int6 -> int7: this is contradiction.; [assump1] & int7 -> int8: the mycophagist is not a practice.; int8 & sent7 -> int9: the mycophagist is not a skim.; sent5 -> int10: if the mycophagist is not a kind of a skim it is a kind of a Polynesian and it does not gross.; int9 & int10 -> int11: the mycophagist is both a Polynesian and not a gross.; int11 & sent1 -> hypothesis; __PROVED__