the fact that the congregationalness occurs and the Bayesianness does not occur does not hold.
sent1: if the packing happens the congregationalness and the non-Bayesianness occurs. sent2: the fact that that both the eyedrop and the mecopterousness occurs is correct is wrong if the commodiousness does not occur. sent3: the packing happens if the burgeoning does not occur. sent4: the commodiousness does not occur. sent5: if the fact that that the eyedrop occurs and the mecopterousness happens is incorrect is not wrong the rainstorm does not occur. sent6: that the rainstorm does not occur yields that the permitting brooklime occurs and the burgeoning does not occur. sent7: if the tearfulness occurs the thermostaticness but not the pupillariness occurs.
sent2 & sent4 -> int1: the fact that the eyedrop and the mecopterousness occurs is false.; sent5 & int1 -> int2: that the rainstorm does not occur is right.; sent6 & int2 -> int3: the permitting brooklime happens but the burgeoning does not occur.; int3 -> int4: the burgeoning does not occur.; sent3 & int4 -> int5: the packing occurs.; sent1 & int5 -> hypothesis;
$hypothesis$ = the fact that the congregationalness occurs and the Bayesianness does not occur does not hold. ; $context$ = sent1: if the packing happens the congregationalness and the non-Bayesianness occurs. sent2: the fact that that both the eyedrop and the mecopterousness occurs is correct is wrong if the commodiousness does not occur. sent3: the packing happens if the burgeoning does not occur. sent4: the commodiousness does not occur. sent5: if the fact that that the eyedrop occurs and the mecopterousness happens is incorrect is not wrong the rainstorm does not occur. sent6: that the rainstorm does not occur yields that the permitting brooklime occurs and the burgeoning does not occur. sent7: if the tearfulness occurs the thermostaticness but not the pupillariness occurs. ; $proof$ = 
sent2 & sent4 -> int1: the fact that the eyedrop and the mecopterousness occurs is false.; sent5 & int1 -> int2: that the rainstorm does not occur is right.; sent6 & int2 -> int3: the permitting brooklime happens but the burgeoning does not occur.; int3 -> int4: the burgeoning does not occur.; sent3 & int4 -> int5: the packing occurs.; sent1 & int5 -> hypothesis; __DISPROVED__