the fact that the bedlamite does not hedge accuracy is correct.
sent1: if the bedlamite does hedge accuracy the evolution is not a bray but it hedges arbor. sent2: if the fact that the bedlamite is aware is not false the evolution does hedge accuracy. sent3: if something is not a weight then that it is not run-on and it does not decompose is incorrect. sent4: if that something restores pasang or it is aware or both is true the MAOI does not decompose. sent5: the MAOI decomposes if that the arbor does not decomposes is true. sent6: something is not a weight or it does not decompose or both if it is run-on. sent7: the fact that something hedges accuracy and it is aware is wrong if it does not restore pasang. sent8: the arbor does not hedge accuracy if there exists something such that it is a kind of a weight and/or it is run-on. sent9: if the arbor is not a weight the MAOI decomposes. sent10: if something is run-on thing that does not decompose then it is not aware. sent11: if that the fact that something is a kind of a elephant's-foot and it is a weight is not wrong is not true then it is not a kind of a weight. sent12: the evolution restores pasang if it is not a bray and hedges arbor. sent13: if there is something such that that it is not run-on and it does not decompose is not true then the bedlamite does not restore pasang. sent14: the arbor is run-on. sent15: that the evolution is a elephant's-foot and a weight is not correct.
void -> assump1: Let's assume that the bedlamite does hedge accuracy.; sent1 & assump1 -> int1: the evolution does not bray but it does hedge arbor.; int1 & sent12 -> int2: the evolution does restore pasang.; int2 -> int3: the evolution does restore pasang and/or is aware.; int3 -> int4: something restores pasang and/or is aware.; int4 & sent4 -> int5: the MAOI does not decompose.; sent6 -> int6: the arbor is not a weight or it does not decompose or both if it is run-on.; int6 & sent14 -> int7: the arbor is not a weight and/or does not decompose.; int7 & sent9 & sent5 -> int8: the MAOI decomposes.; int5 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis;
the bedlamite does hedge accuracy.
sent10 -> int10: the evolution is not aware if it is a kind of run-on thing that does not decompose.;
$hypothesis$ = the fact that the bedlamite does not hedge accuracy is correct. ; $context$ = sent1: if the bedlamite does hedge accuracy the evolution is not a bray but it hedges arbor. sent2: if the fact that the bedlamite is aware is not false the evolution does hedge accuracy. sent3: if something is not a weight then that it is not run-on and it does not decompose is incorrect. sent4: if that something restores pasang or it is aware or both is true the MAOI does not decompose. sent5: the MAOI decomposes if that the arbor does not decomposes is true. sent6: something is not a weight or it does not decompose or both if it is run-on. sent7: the fact that something hedges accuracy and it is aware is wrong if it does not restore pasang. sent8: the arbor does not hedge accuracy if there exists something such that it is a kind of a weight and/or it is run-on. sent9: if the arbor is not a weight the MAOI decomposes. sent10: if something is run-on thing that does not decompose then it is not aware. sent11: if that the fact that something is a kind of a elephant's-foot and it is a weight is not wrong is not true then it is not a kind of a weight. sent12: the evolution restores pasang if it is not a bray and hedges arbor. sent13: if there is something such that that it is not run-on and it does not decompose is not true then the bedlamite does not restore pasang. sent14: the arbor is run-on. sent15: that the evolution is a elephant's-foot and a weight is not correct. ; $proof$ = 
void -> assump1: Let's assume that the bedlamite does hedge accuracy.; sent1 & assump1 -> int1: the evolution does not bray but it does hedge arbor.; int1 & sent12 -> int2: the evolution does restore pasang.; int2 -> int3: the evolution does restore pasang and/or is aware.; int3 -> int4: something restores pasang and/or is aware.; int4 & sent4 -> int5: the MAOI does not decompose.; sent6 -> int6: the arbor is not a weight or it does not decompose or both if it is run-on.; int6 & sent14 -> int7: the arbor is not a weight and/or does not decompose.; int7 & sent9 & sent5 -> int8: the MAOI decomposes.; int5 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis; __PROVED__