if the disposal is not intrasentential it does not rerun reliquary and it is ineffective.
sent1: something does not foul if it does not hoard. sent2: the disposal is ineffective. sent3: the disposal is intrasentential if it reruns reliquary. sent4: the disposal does not panic areca. sent5: the asp is ineffective if the fact that the disposal is not intrasentential and reruns reliquary is false. sent6: the sharp is not intrasentential. sent7: if the brogan is not a kind of a foul the disposal is a kind of a foul. sent8: if that the Wintun is inconvenient is true then it is not a cytology and it is intrasentential. sent9: the disposal does not rerun reliquary but it is ineffective if it is intrasentential. sent10: the fact that something is not intrasentential and does rerun reliquary is incorrect if it fouls.
void -> assump2: Let's assume that the disposal reruns reliquary.; sent3 & assump2 -> int1: the fact that the disposal is intrasentential is correct.; void -> assump1: Let's assume that the disposal is not intrasentential.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the disposal does not rerun reliquary.; int3 & sent2 -> int4: the disposal does not rerun reliquary but it is ineffective.; [assump1] & int4 -> hypothesis;
the asp is ineffective.
sent10 -> int5: if the disposal fouls the fact that it is not intrasentential and it reruns reliquary is incorrect.; sent1 -> int6: if the brogan is not a hoard then it is non-foul.;
$hypothesis$ = if the disposal is not intrasentential it does not rerun reliquary and it is ineffective. ; $context$ = sent1: something does not foul if it does not hoard. sent2: the disposal is ineffective. sent3: the disposal is intrasentential if it reruns reliquary. sent4: the disposal does not panic areca. sent5: the asp is ineffective if the fact that the disposal is not intrasentential and reruns reliquary is false. sent6: the sharp is not intrasentential. sent7: if the brogan is not a kind of a foul the disposal is a kind of a foul. sent8: if that the Wintun is inconvenient is true then it is not a cytology and it is intrasentential. sent9: the disposal does not rerun reliquary but it is ineffective if it is intrasentential. sent10: the fact that something is not intrasentential and does rerun reliquary is incorrect if it fouls. ; $proof$ = 
void -> assump2: Let's assume that the disposal reruns reliquary.; sent3 & assump2 -> int1: the fact that the disposal is intrasentential is correct.; void -> assump1: Let's assume that the disposal is not intrasentential.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the disposal does not rerun reliquary.; int3 & sent2 -> int4: the disposal does not rerun reliquary but it is ineffective.; [assump1] & int4 -> hypothesis; __PROVED__