the stem is blinded.
sent1: that the stem is a NRC and does not arbitrage is wrong if there is something such that it is not a NRC. sent2: the fact that something is not ammonitic but non-blinded does not hold if that it is not asphaltic is correct. sent3: if something is not a NRC then the fact that the trave is a NRC but it is not a Diervilla is not true. sent4: something is not a NRC. sent5: the stem is not a kind of a Diervilla. sent6: if the fact that something is a NRC and is not a kind of an arbitrage is wrong it is not asphaltic. sent7: that the trave is not a graphology but it is approachable is not true if it is not a kind of a Diervilla. sent8: the stem is not ammonitic if the fact that the trave is not a graphology but approachable is not right. sent9: if something is an arbitrage then it is not asphaltic. sent10: that the trave is not a Diervilla hold. sent11: if the trave is ammonitic then the stem is not a Diervilla.
sent7 & sent10 -> int1: that the trave is not a kind of a graphology and is approachable does not hold.; int1 & sent8 -> int2: the stem is not ammonitic.; void -> assump1: Let's assume that the stem is not blinded.; int2 & assump1 -> int3: the stem is a kind of non-ammonitic thing that is non-blinded.; sent2 -> int4: if the stem is non-asphaltic then the fact that it is not ammonitic and it is not a blinded is not right.; sent6 -> int5: if the fact that the stem is a NRC but it is not a kind of an arbitrage is not true then that it is not asphaltic hold.; sent4 & sent1 -> int6: the fact that the stem is a kind of a NRC but it does not arbitrage is false.; int5 & int6 -> int7: the stem is not asphaltic.; int4 & int7 -> int8: the fact that that the stem is not ammonitic and is non-blinded is incorrect is right.; int3 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis;
$hypothesis$ = the stem is blinded. ; $context$ = sent1: that the stem is a NRC and does not arbitrage is wrong if there is something such that it is not a NRC. sent2: the fact that something is not ammonitic but non-blinded does not hold if that it is not asphaltic is correct. sent3: if something is not a NRC then the fact that the trave is a NRC but it is not a Diervilla is not true. sent4: something is not a NRC. sent5: the stem is not a kind of a Diervilla. sent6: if the fact that something is a NRC and is not a kind of an arbitrage is wrong it is not asphaltic. sent7: that the trave is not a graphology but it is approachable is not true if it is not a kind of a Diervilla. sent8: the stem is not ammonitic if the fact that the trave is not a graphology but approachable is not right. sent9: if something is an arbitrage then it is not asphaltic. sent10: that the trave is not a Diervilla hold. sent11: if the trave is ammonitic then the stem is not a Diervilla. ; $proof$ = 
sent7 & sent10 -> int1: that the trave is not a kind of a graphology and is approachable does not hold.; int1 & sent8 -> int2: the stem is not ammonitic.; void -> assump1: Let's assume that the stem is not blinded.; int2 & assump1 -> int3: the stem is a kind of non-ammonitic thing that is non-blinded.; sent2 -> int4: if the stem is non-asphaltic then the fact that it is not ammonitic and it is not a blinded is not right.; sent6 -> int5: if the fact that the stem is a NRC but it is not a kind of an arbitrage is not true then that it is not asphaltic hold.; sent4 & sent1 -> int6: the fact that the stem is a kind of a NRC but it does not arbitrage is false.; int5 & int6 -> int7: the stem is not asphaltic.; int4 & int7 -> int8: the fact that that the stem is not ammonitic and is non-blinded is incorrect is right.; int3 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis; __PROVED__