the lappet trivializes epicycle.
sent1: the shaver is inefficient. sent2: the shaver is not a Wollaston. sent3: if something does not knit Anisoptera that it is not inefficient and/or it is not Corsican is wrong. sent4: something trivializes epicycle if that it does not signify and it does not trivializes epicycle is wrong. sent5: the shaver does not knit Anisoptera if that the nemertean is a kind of a Californian and/or does not burl is false. sent6: something is not Californian if it does signify. sent7: if something is not a Californian then the fact that it does not signify and does not trivialize epicycle is incorrect. sent8: something is not inefficient or is not Corsican or both if it is not a Wollaston. sent9: that something is a Californian or is not a burl or both does not hold if that it does signify is correct. sent10: if the lappet does trivialize epicycle then the nemertean signifies. sent11: something is not Californian if it does not knit Anisoptera or is not a Californian or both.
void -> assump1: Let's assume that the lappet trivializes epicycle.; sent10 & assump1 -> int1: the nemertean signifies.; sent9 -> int2: if the nemertean does signify the fact that the fact that it is Californian and/or it does not burl hold is incorrect.; int1 & int2 -> int3: that the nemertean is a kind of a Californian or it is not a burl or both is incorrect.; int3 & sent5 -> int4: the shaver does not knit Anisoptera.; sent3 -> int5: if the shaver does not knit Anisoptera that it is efficient and/or non-Corsican is false.; int4 & int5 -> int6: the fact that the shaver is not inefficient and/or is not Corsican is false.; sent8 -> int7: if the shaver is not a Wollaston then it is not inefficient and/or is not Corsican.; int7 & sent2 -> int8: the shaver is not inefficient or it is not Corsican or both.; int6 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis;
the lappet trivializes epicycle.
sent4 -> int10: the lappet does trivialize epicycle if the fact that it does not signify and does not trivialize epicycle is not correct.; sent7 -> int11: if the lappet is not Californian that it does not signify and does not trivialize epicycle does not hold.; sent11 -> int12: if the lappet does not knit Anisoptera and/or it is not Californian then it is not a Californian.;
$hypothesis$ = the lappet trivializes epicycle. ; $context$ = sent1: the shaver is inefficient. sent2: the shaver is not a Wollaston. sent3: if something does not knit Anisoptera that it is not inefficient and/or it is not Corsican is wrong. sent4: something trivializes epicycle if that it does not signify and it does not trivializes epicycle is wrong. sent5: the shaver does not knit Anisoptera if that the nemertean is a kind of a Californian and/or does not burl is false. sent6: something is not Californian if it does signify. sent7: if something is not a Californian then the fact that it does not signify and does not trivialize epicycle is incorrect. sent8: something is not inefficient or is not Corsican or both if it is not a Wollaston. sent9: that something is a Californian or is not a burl or both does not hold if that it does signify is correct. sent10: if the lappet does trivialize epicycle then the nemertean signifies. sent11: something is not Californian if it does not knit Anisoptera or is not a Californian or both. ; $proof$ = 
void -> assump1: Let's assume that the lappet trivializes epicycle.; sent10 & assump1 -> int1: the nemertean signifies.; sent9 -> int2: if the nemertean does signify the fact that the fact that it is Californian and/or it does not burl hold is incorrect.; int1 & int2 -> int3: that the nemertean is a kind of a Californian or it is not a burl or both is incorrect.; int3 & sent5 -> int4: the shaver does not knit Anisoptera.; sent3 -> int5: if the shaver does not knit Anisoptera that it is efficient and/or non-Corsican is false.; int4 & int5 -> int6: the fact that the shaver is not inefficient and/or is not Corsican is false.; sent8 -> int7: if the shaver is not a Wollaston then it is not inefficient and/or is not Corsican.; int7 & sent2 -> int8: the shaver is not inefficient or it is not Corsican or both.; int6 & int8 -> int9: this is contradiction.; [assump1] & int9 -> hypothesis; __DISPROVED__