ok here’s another one…a lil more complex, but still not 2 bad…just not sure bout the end though…it’s a long one…if anyone has another way of proving this, plz let me know…
symbols that i’m usin here, “-” represents “Not/Negation” ; “–” represents “Double Negation”; “==” represents “Is Equivalent to” ; “=” is “equals” ; “:=” represents “is substitued by” ; “!==” represents “is not equivalent to”…so here it is:
Theorem: ( ( p !== q ) !== r ) == ( p !== ( q !== r ) )
on ( ( p !== q ) !== r ) from the theorem, applying Definition of !==: ( p !== q ) == - ( p == q ) where p:= ( p !== q ) and q:= r ;and Leibniz on the rest:
= - ( ( p !== q ) == r ) == ( p !== ( q !== r ) )
on ( p !== ( q !== r ) ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) where q:= ( q !== r ) ;and Leibniz on the rest:
= - ( ( p !== q ) == r ) == - ( p == ( q !== r ) )
on ( p !== q ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) ; and Leibniz on the rest:
= - ( - ( p == q ) == r ) == - ( p == ( q !== r ) )
on ( q !== r ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) where p:= q, q:= r ; and Leibniz on the rest:
= - ( - ( p == q ) == r ) == - ( p == - ( q == r ) )
on - ( p == q ) from the above statement, applying Distributivity of - over ==: - ( p == q ) == - p == q ; and Leibniz on rest:
= - ( - p == q == r ) == - ( p == - ( q == r ) )
on - ( q == r ) from the above statement, applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= q, q:= r; and Leibniz on rest:
= - ( - p == q == r ) == - ( p == - q == r )
on ( p == - q ) from the above statement, applying Symmetry of ==: p == q == q == p where q:= - q; and Leibniz on rest:
= - ( - p == q == r ) == - ( - q == p == r )
on - ( - p == q == r ), applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= - p, q:= q == r; and Leibniz on rest:
= – p == q == r == - ( - q == p == r )
on - ( - q == p == r ), applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= - q, q:= p == r; and Leibniz on rest:
= – p == q == r == – q == p == r
From Previous Theorem, – p == p, therefore:
= p == q == r == – q == p == r
on – q, applying Theorem: – p == p where p == q; and Leibniz on rest:
= p == q == r == q == p == r
on ( q == p ) from the above statement, applying Symmetry of ==: p == q == q == p; and Leibniz on rest:
= p == q == r == p == q == r
Now here is what i don’t know…
Using the Axiom of Associativity of ==: ( p == ( q == r ) ) == ( ( p == q ) == r ) -----> Is it a valid step??
or
Using Theorem of Relexivity of ==: p == p where p:= p == q == r -----> this should be valid for sure
phew
comeon…where are all the math experts here?? not that an expert is needed for this proof
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