here's the theorem that i'm tryin to prove...it's a basic one, i just wanna be sure that i am doin this right...symbols that i'm usin here, "-" represents "NOT" and "==" represents "IS EQUIVALENT TO"...and of course, "=" is equals, and ":=" represents "is substitued by"...so here it is

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Theorem: -- p == q == p == --q

Using Distribution of - over ==: - (p == q) == -p == q, for "--p == q" from the theorem; and using Leibniz for the rest of the theorem to carry it down as it is:

= - (- (p == q)) == p == --q

Using Symmetry of == on "p == --q" from the last line: p == q == q == p, where q:= --q; and leibneiz on the rest:

= - (- (p == q)) == --q == p

for "--q == p" from the last line, using Distribution of - over ==: - (p == q) == -p == q, where p:= q, q:= p; and using Leibniz for the rest to carry it down as it is:

= - (- (p == q)) == - (- (p == q))

Reflexivity of == states: p == p; here p:= -( - (p ==q)), therefore intitial theorm is true.

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**phew**

can someone look over it, c if it's correct? and can someone think of any other ways to prove the theorem??

thanx in advance

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Theorem: -- p == q == p == --q

Using Distribution of - over ==: - (p == q) == -p == q, for "--p == q" from the theorem; and using Leibniz for the rest of the theorem to carry it down as it is:

= - (- (p == q)) == p == --q

Using Symmetry of == on "p == --q" from the last line: p == q == q == p, where q:= --q; and leibneiz on the rest:

= - (- (p == q)) == --q == p

for "--q == p" from the last line, using Distribution of - over ==: - (p == q) == -p == q, where p:= q, q:= p; and using Leibniz for the rest to carry it down as it is:

= - (- (p == q)) == - (- (p == q))

Reflexivity of == states: p == p; here p:= -( - (p ==q)), therefore intitial theorm is true.

--------------------------------------------

**phew**

can someone look over it, c if it's correct? and can someone think of any other ways to prove the theorem??

thanx in advance

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**If I'm dreaming, never let me wake. If I'm awake, never let me sleep.**
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