the fact that the redhorse is a spruce but it is not a discretion does not hold.
sent1: if that the redhorse is a kind of a horsemanship is right then that it is a centrifugation that is not unicameral hold. sent2: a spruce that is not a discretion confuses fixer-upper. sent3: the bullion confuses fixer-upper. sent4: if the redhorse is a kind of a centrifugation that is clockwise then it does confuse hasp. sent5: the redhorse is clockwise. sent6: that the fact that the cote does confuse fixer-upper and it is a Bryan is wrong is true. sent7: the fact that something is a kind of a centrifugation and is not unicameral is false if it does confuse fixer-upper. sent8: if the redhorse is a rifampin and it is thermodynamic it is unicameral. sent9: that the omnibus is an affiliate is correct if it is Hebridean. sent10: if something is unicameral it is a kind of a spruce that is not a discretion. sent11: the bottle-grass repines selling and is not unicameral. sent12: the redhorse is a horsemanship and is classificatory. sent13: that the redhorse is a centrifugation is not incorrect.
sent2 -> int1: if the redhorse is a spruce and is not a discretion it does confuse fixer-upper.; void -> assump1: Let's assume that the redhorse is a kind of a spruce but it is not a discretion.; int1 & assump1 -> int2: the redhorse does confuse fixer-upper.; sent7 -> int3: if the redhorse confuses fixer-upper the fact that it is a centrifugation and not unicameral is not correct.; int2 & int3 -> int4: the fact that the redhorse is a centrifugation but it is not unicameral is wrong.; sent12 -> int5: the redhorse is a kind of a horsemanship.; sent1 & int5 -> int6: the redhorse is a centrifugation and not unicameral.; int4 & int6 -> int7: this is contradiction.; [assump1] & int7 -> hypothesis;
the redhorse does spruce but it is not a discretion.
sent10 -> int8: the redhorse does spruce but it is not a discretion if it is unicameral.;
$hypothesis$ = the fact that the redhorse is a spruce but it is not a discretion does not hold. ; $context$ = sent1: if that the redhorse is a kind of a horsemanship is right then that it is a centrifugation that is not unicameral hold. sent2: a spruce that is not a discretion confuses fixer-upper. sent3: the bullion confuses fixer-upper. sent4: if the redhorse is a kind of a centrifugation that is clockwise then it does confuse hasp. sent5: the redhorse is clockwise. sent6: that the fact that the cote does confuse fixer-upper and it is a Bryan is wrong is true. sent7: the fact that something is a kind of a centrifugation and is not unicameral is false if it does confuse fixer-upper. sent8: if the redhorse is a rifampin and it is thermodynamic it is unicameral. sent9: that the omnibus is an affiliate is correct if it is Hebridean. sent10: if something is unicameral it is a kind of a spruce that is not a discretion. sent11: the bottle-grass repines selling and is not unicameral. sent12: the redhorse is a horsemanship and is classificatory. sent13: that the redhorse is a centrifugation is not incorrect. ; $proof$ = 
sent2 -> int1: if the redhorse is a spruce and is not a discretion it does confuse fixer-upper.; void -> assump1: Let's assume that the redhorse is a kind of a spruce but it is not a discretion.; int1 & assump1 -> int2: the redhorse does confuse fixer-upper.; sent7 -> int3: if the redhorse confuses fixer-upper the fact that it is a centrifugation and not unicameral is not correct.; int2 & int3 -> int4: the fact that the redhorse is a centrifugation but it is not unicameral is wrong.; sent12 -> int5: the redhorse is a kind of a horsemanship.; sent1 & int5 -> int6: the redhorse is a centrifugation and not unicameral.; int4 & int6 -> int7: this is contradiction.; [assump1] & int7 -> hypothesis; __PROVED__