the fact that if the passage is retinal then the fact that the usance is not a kind of a fancy and it is non-photomechanical does not hold is wrong.
sent1: if the fact that the usance is a fancy hold that the passage is not retinal and not photomechanical is not right. sent2: something that fructifies is a kind of non-fancy thing that is not a kind of a bachelorhood. sent3: the usance is both not a fancy and not photomechanical if the passage is retinal. sent4: the passage is not retinal if the usance is not a kind of a fancy and it is non-photomechanical. sent5: the passage is not retinal if the usance is fancy thing that is not photomechanical. sent6: if the fact that the passage is retinal is not wrong the fact that the fact that the usance is both non-fancy and photomechanical is false is right. sent7: if the fact that the usance is retinal hold the passage fructifies. sent8: the usance is not retinal and it is not photomechanical. sent9: the passage is non-retinal if the usance is not a fancy but it is photomechanical. sent10: if the usance is non-fancy and is not photomechanical then the passage is retinal. sent11: the usance is non-photomechanical. sent12: if the fact that the usance is not retinal and is non-fancy is right the passage is not photomechanical. sent13: something is both retinal and clinker-built if the fact that it does not camber hospital is not incorrect. sent14: that the usance does fancy but it is not photomechanical is not right if the passage is retinal. sent15: the passage does not maunder and it is not mindful. sent16: that the usance is both not a fancy and not non-photomechanical is false.
void -> assump2: Let's assume that the usance is non-fancy thing that is not photomechanical.; sent4 & assump2 -> int1: the passage is not retinal.; void -> assump1: Let's assume that the passage is retinal.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the fact that the usance is non-fancy and it is not photomechanical is not true.; [assump1] & int3 -> hypothesis;
the passage is not a fancy and it is not a bachelorhood.
sent2 -> int4: the passage is a kind of non-fancy thing that is not a bachelorhood if it does fructify.; sent13 -> int5: if the usance does not camber hospital it is retinal and clinker-built.;
$hypothesis$ = the fact that if the passage is retinal then the fact that the usance is not a kind of a fancy and it is non-photomechanical does not hold is wrong. ; $context$ = sent1: if the fact that the usance is a fancy hold that the passage is not retinal and not photomechanical is not right. sent2: something that fructifies is a kind of non-fancy thing that is not a kind of a bachelorhood. sent3: the usance is both not a fancy and not photomechanical if the passage is retinal. sent4: the passage is not retinal if the usance is not a kind of a fancy and it is non-photomechanical. sent5: the passage is not retinal if the usance is fancy thing that is not photomechanical. sent6: if the fact that the passage is retinal is not wrong the fact that the fact that the usance is both non-fancy and photomechanical is false is right. sent7: if the fact that the usance is retinal hold the passage fructifies. sent8: the usance is not retinal and it is not photomechanical. sent9: the passage is non-retinal if the usance is not a fancy but it is photomechanical. sent10: if the usance is non-fancy and is not photomechanical then the passage is retinal. sent11: the usance is non-photomechanical. sent12: if the fact that the usance is not retinal and is non-fancy is right the passage is not photomechanical. sent13: something is both retinal and clinker-built if the fact that it does not camber hospital is not incorrect. sent14: that the usance does fancy but it is not photomechanical is not right if the passage is retinal. sent15: the passage does not maunder and it is not mindful. sent16: that the usance is both not a fancy and not non-photomechanical is false. ; $proof$ = 
void -> assump2: Let's assume that the usance is non-fancy thing that is not photomechanical.; sent4 & assump2 -> int1: the passage is not retinal.; void -> assump1: Let's assume that the passage is retinal.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the fact that the usance is non-fancy and it is not photomechanical is not true.; [assump1] & int3 -> hypothesis; __DISPROVED__