the sauce-alone does spread-eagle if the pontoon is not a Goldwyn but it is a consultancy.
sent1: the fact that the pontoon is a Goldwyn and it is a consultancy is false if the sauce-alone does not spread-eagle. sent2: the pontoon is not a kind of a Goldwyn but it is a consultancy if the sauce-alone does not spread-eagle. sent3: if the fact that the sauce-alone does not spread-eagle is true that that the pontoon is both not a Goldwyn and a consultancy is correct is not true. sent4: the fact that the pontoon is a Goldwyn and spread-eagles does not hold. sent5: the pontoon is not a Goldwyn. sent6: if the sauce-alone is not a consultancy the pontoon is not a Goldwyn but it does spread-eagle. sent7: the sauce-alone is not a kind of a Goldwyn. sent8: the tympanum is a Goldwyn. sent9: if the pontoon is a Goldwyn and it is a consultancy then the sauce-alone spread-eagles. sent10: that the sauce-alone is a kind of a consultancy and does spread-eagle is false. sent11: the fact that the pontoon is a Goldwyn and a consultancy does not hold. sent12: that the pontoon is not a kind of a Goldwyn but it is a consultancy is incorrect if the sauce-alone does spread-eagle. sent13: the tamp is a Goldwyn.
void -> assump2: Let's assume that the sauce-alone does not spread-eagle.; sent3 & assump2 -> int1: that the pontoon is not a Goldwyn but it is a consultancy does not hold.; void -> assump1: Let's assume that the pontoon is both not a Goldwyn and a consultancy.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the sauce-alone spread-eagles.; [assump1] & int3 -> hypothesis;
$hypothesis$ = the sauce-alone does spread-eagle if the pontoon is not a Goldwyn but it is a consultancy. ; $context$ = sent1: the fact that the pontoon is a Goldwyn and it is a consultancy is false if the sauce-alone does not spread-eagle. sent2: the pontoon is not a kind of a Goldwyn but it is a consultancy if the sauce-alone does not spread-eagle. sent3: if the fact that the sauce-alone does not spread-eagle is true that that the pontoon is both not a Goldwyn and a consultancy is correct is not true. sent4: the fact that the pontoon is a Goldwyn and spread-eagles does not hold. sent5: the pontoon is not a Goldwyn. sent6: if the sauce-alone is not a consultancy the pontoon is not a Goldwyn but it does spread-eagle. sent7: the sauce-alone is not a kind of a Goldwyn. sent8: the tympanum is a Goldwyn. sent9: if the pontoon is a Goldwyn and it is a consultancy then the sauce-alone spread-eagles. sent10: that the sauce-alone is a kind of a consultancy and does spread-eagle is false. sent11: the fact that the pontoon is a Goldwyn and a consultancy does not hold. sent12: that the pontoon is not a kind of a Goldwyn but it is a consultancy is incorrect if the sauce-alone does spread-eagle. sent13: the tamp is a Goldwyn. ; $proof$ = 
void -> assump2: Let's assume that the sauce-alone does not spread-eagle.; sent3 & assump2 -> int1: that the pontoon is not a Goldwyn but it is a consultancy does not hold.; void -> assump1: Let's assume that the pontoon is both not a Goldwyn and a consultancy.; int1 & assump1 -> int2: this is contradiction.; [assump2] & int2 -> int3: the sauce-alone spread-eagles.; [assump1] & int3 -> hypothesis; __PROVED__