if the invite is intentional then the boxfish is not a desalination.
sent1: the fact that the invite does not adjoin river is right. sent2: if something is a kind of a desalination it does not adjoin river. sent3: if the invite is a desalination then the classic is intentional. sent4: if the boxfish does not adjoin river the invite does not adjoin river. sent5: if the invite does not adjoin river and it is not Merovingian then it is intentional. sent6: the invite is not intentional if it does adjoin river and it is not Merovingian. sent7: the invite is not intentional if it does not adjoin river and it is not a kind of a Merovingian. sent8: the invite does not adjoin river and it is non-Merovingian if the boxfish does not adjoin river.
sent2 -> int1: if the boxfish is a desalination it does not adjoin river.; void -> assump2: Let's assume that the boxfish is a desalination.; int1 & assump2 -> int2: the boxfish does not adjoin river.; int2 & sent8 -> int3: the invite does not adjoin river and is not a Merovingian.; int3 & sent7 -> int4: the invite is not intentional.; void -> assump1: Let's assume that the invite is intentional.; int4 & assump1 -> int5: this is contradiction.; [assump2] & int5 -> int6: the boxfish is not a desalination.; [assump1] & int6 -> hypothesis;
the classic is intentional.
$hypothesis$ = if the invite is intentional then the boxfish is not a desalination. ; $context$ = sent1: the fact that the invite does not adjoin river is right. sent2: if something is a kind of a desalination it does not adjoin river. sent3: if the invite is a desalination then the classic is intentional. sent4: if the boxfish does not adjoin river the invite does not adjoin river. sent5: if the invite does not adjoin river and it is not Merovingian then it is intentional. sent6: the invite is not intentional if it does adjoin river and it is not Merovingian. sent7: the invite is not intentional if it does not adjoin river and it is not a kind of a Merovingian. sent8: the invite does not adjoin river and it is non-Merovingian if the boxfish does not adjoin river. ; $proof$ = 
sent2 -> int1: if the boxfish is a desalination it does not adjoin river.; void -> assump2: Let's assume that the boxfish is a desalination.; int1 & assump2 -> int2: the boxfish does not adjoin river.; int2 & sent8 -> int3: the invite does not adjoin river and is not a Merovingian.; int3 & sent7 -> int4: the invite is not intentional.; void -> assump1: Let's assume that the invite is intentional.; int4 & assump1 -> int5: this is contradiction.; [assump2] & int5 -> int6: the boxfish is not a desalination.; [assump1] & int6 -> hypothesis; __PROVED__