the adenoid overturns glassmaker.
sent1: if something is urinary then it is not a row and does group reedbird. sent2: the fact that the Hebraist is not a workbook is correct if the fact that it is a kind of a haze is true. sent3: either the Hebraist is a kind of a haze or it does not barbarize tide or both. sent4: there is something such that it is a workbook. sent5: something does not overturn glassmaker if that it is non-Burmese thing that is not a workbook is false. sent6: if something does not row the fact that it is not Burmese and it is not a workbook does not hold. sent7: the Hebraist does row and it is Burmese if the glassmaker is not a workbook. sent8: if the Hebraist is a haze or it does not barbarize tide or both then it is not a workbook. sent9: the adenoid is a Burmese if that the glassmaker does overturn glassmaker is not incorrect. sent10: if there exists something such that it is not a workbook then the glassmaker is a Burmese. sent11: if something does not group reedbird then it does not overturn glassmaker and it is not a workbook. sent12: something that is a kind of a Burmese is a row.
sent8 & sent3 -> int1: the Hebraist is not a workbook.; int1 -> int2: something is not a kind of a workbook.; int2 & sent10 -> int3: the glassmaker is a kind of a Burmese.; sent12 -> int4: if the glassmaker is a Burmese it is a row.; int3 & int4 -> int5: the glassmaker is a kind of a row.;
the Hebraist is a row.
sent11 -> int6: the glassmaker does not overturn glassmaker and it is not a kind of a workbook if it does not group reedbird.;
$hypothesis$ = the adenoid overturns glassmaker. ; $context$ = sent1: if something is urinary then it is not a row and does group reedbird. sent2: the fact that the Hebraist is not a workbook is correct if the fact that it is a kind of a haze is true. sent3: either the Hebraist is a kind of a haze or it does not barbarize tide or both. sent4: there is something such that it is a workbook. sent5: something does not overturn glassmaker if that it is non-Burmese thing that is not a workbook is false. sent6: if something does not row the fact that it is not Burmese and it is not a workbook does not hold. sent7: the Hebraist does row and it is Burmese if the glassmaker is not a workbook. sent8: if the Hebraist is a haze or it does not barbarize tide or both then it is not a workbook. sent9: the adenoid is a Burmese if that the glassmaker does overturn glassmaker is not incorrect. sent10: if there exists something such that it is not a workbook then the glassmaker is a Burmese. sent11: if something does not group reedbird then it does not overturn glassmaker and it is not a workbook. sent12: something that is a kind of a Burmese is a row. ; $proof$ = 
sent8 & sent3 -> int1: the Hebraist is not a workbook.; int1 -> int2: something is not a kind of a workbook.; int2 & sent10 -> int3: the glassmaker is a kind of a Burmese.; sent12 -> int4: if the glassmaker is a Burmese it is a row.; int3 & int4 -> int5: the glassmaker is a kind of a row.; __UNKNOWN__