that the spit does not occur results in that the typhoon does not occur but the corruptedness happens.
sent1: the typhoon does not occur and the spit does not occur if the corrupting does not occur. sent2: the queuing horniness happens if the typhoon does not occur and the spitting does not occur. sent3: the fact that the blinkering tracheid but not the farming dainty happens is not right. sent4: the stream does not occur but the brisantness occurs. sent5: that the blinkering tracheid does not occur and the farming dainty does not occur is not correct. sent6: if that the blinkering tracheid does not occur and the farming dainty does not occur is not true then the feature does not occur. sent7: the spit happens if the typhoon happens.
void -> assump1: Let's assume that the typhoon happens.; sent7 & assump1 -> int1: that the spitting happens is not wrong.; void -> assump2: Let's assume that the spit does not occur.; int1 & assump2 -> int2: this is contradiction.; [assump1] & int2 -> int3: the typhoon does not occur.; sent6 & sent5 -> int4: the feature does not occur.;
the queuing horniness happens.
$hypothesis$ = that the spit does not occur results in that the typhoon does not occur but the corruptedness happens. ; $context$ = sent1: the typhoon does not occur and the spit does not occur if the corrupting does not occur. sent2: the queuing horniness happens if the typhoon does not occur and the spitting does not occur. sent3: the fact that the blinkering tracheid but not the farming dainty happens is not right. sent4: the stream does not occur but the brisantness occurs. sent5: that the blinkering tracheid does not occur and the farming dainty does not occur is not correct. sent6: if that the blinkering tracheid does not occur and the farming dainty does not occur is not true then the feature does not occur. sent7: the spit happens if the typhoon happens. ; $proof$ = 
void -> assump1: Let's assume that the typhoon happens.; sent7 & assump1 -> int1: that the spitting happens is not wrong.; void -> assump2: Let's assume that the spit does not occur.; int1 & assump2 -> int2: this is contradiction.; [assump1] & int2 -> int3: the typhoon does not occur.; sent6 & sent5 -> int4: the feature does not occur.; __UNKNOWN__