the candlewood expenses compromising and it does misbelieve.
sent1: if the quack does shampoo phial it is attritional. sent2: the quack does not concern lazuli and shampoos phial if the dray is attritional. sent3: if something does not shampoo phial then the fact that it expenses compromising and does misbelieve does not hold. sent4: the quack is not attritional. sent5: the liveryman is a kind of a Gibraltar and reopens Dail. sent6: everything does expense veery. sent7: the candlewood does not concern lazuli but it does shampoo phial if the quack does not shampoo phial. sent8: the candlewood expenses veery. sent9: the candlewood does not shampoo phial if the quack does not concern lazuli but it does shampoo phial. sent10: everything expenses veery and it misbelieves. sent11: the candlewood is voltaic and does expense liveryman. sent12: the candlewood does expense solute and is a kind of an underbrush. sent13: the candlewood is not a kind of a camphor if it does concern lazuli and does shampoo phial. sent14: the candlewood does not concern lazuli if the quack does not shampoo phial. sent15: the candlewood expenses compromising if it is not a camphor. sent16: the quack is Tunisian if it is not a Samoan. sent17: the fact that the candlewood does not concern lazuli is true.
void -> assump1: Let's assume that that the quack does shampoo phial is not false.; sent1 & assump1 -> int1: the quack is attritional.; int1 & sent4 -> int2: this is contradiction.; [assump1] & int2 -> int3: the quack does not shampoo phial.; int3 & sent7 -> int4: the candlewood does not concern lazuli but it shampoos phial.; sent10 -> int5: the candlewood expenses veery and misbelieves.; int5 -> int6: the candlewood does misbelieve.;
the fact that the candlewood does expense compromising and misbelieves is not correct.
sent3 -> int7: that the candlewood does expense compromising and does misbelieve is false if it does not shampoo phial.;
$hypothesis$ = the candlewood expenses compromising and it does misbelieve. ; $context$ = sent1: if the quack does shampoo phial it is attritional. sent2: the quack does not concern lazuli and shampoos phial if the dray is attritional. sent3: if something does not shampoo phial then the fact that it expenses compromising and does misbelieve does not hold. sent4: the quack is not attritional. sent5: the liveryman is a kind of a Gibraltar and reopens Dail. sent6: everything does expense veery. sent7: the candlewood does not concern lazuli but it does shampoo phial if the quack does not shampoo phial. sent8: the candlewood expenses veery. sent9: the candlewood does not shampoo phial if the quack does not concern lazuli but it does shampoo phial. sent10: everything expenses veery and it misbelieves. sent11: the candlewood is voltaic and does expense liveryman. sent12: the candlewood does expense solute and is a kind of an underbrush. sent13: the candlewood is not a kind of a camphor if it does concern lazuli and does shampoo phial. sent14: the candlewood does not concern lazuli if the quack does not shampoo phial. sent15: the candlewood expenses compromising if it is not a camphor. sent16: the quack is Tunisian if it is not a Samoan. sent17: the fact that the candlewood does not concern lazuli is true. ; $proof$ = 
void -> assump1: Let's assume that that the quack does shampoo phial is not false.; sent1 & assump1 -> int1: the quack is attritional.; int1 & sent4 -> int2: this is contradiction.; [assump1] & int2 -> int3: the quack does not shampoo phial.; int3 & sent7 -> int4: the candlewood does not concern lazuli but it shampoos phial.; sent10 -> int5: the candlewood expenses veery and misbelieves.; int5 -> int6: the candlewood does misbelieve.; __UNKNOWN__