can Not logic move into this custom logic

Question

(Observeration, Hypothesis) = (then,if)

I want to use all logic with just a custom logic MP(A,B), simply notation as (A,B)

and convert all basic logics into above definition

A -> B = (B,A)  --- implication
(not (not A)) v B = (B,not A) -- Disj
not (not (A ^ B)) = not( (not A) v (not B) ) = not(not B, A) -- Conj

when i meet Disj and Conj

Is not(not B, A) = (B, not A) ?

if so, i am confused as it conclude Disj = Conj ?!

the reason i ask this is that Not logic make pattern not match

i have thought to make not(Prop("Go")) to become Prop("not Go") if not logic can move into proposition

if not logic has distributivity

i design this

(Observeration, Hypothesis) = (then,if)

because convenient of calculation

however, i do not understand not logic applied in not(Observeration, Hypothesis) if it can not move into bracket to become (not Observeration, not Hypothesis)

or

should it not(Observeration, Hypothesis) = (Observeration, not Hypothesis) ?correct?

amWhy · Answer 1 · 2014-03-12T14:42:49.177

Is not(not B, A) = (B, not A) ?

Now, $$\lnot(\lnot b, a) \iff \lnot(a \rightarrow \lnot b) \iff \lnot(\lnot a \lor \lnot b) \iff a \land b$$

$$(b, \lnot a) \iff \lnot a \rightarrow b \iff a \lor b$$

You decide: $$\text{Is}\;a \land b \overset{?}{\equiv} a \lor b\;?$$ (Consider the case when $a$ is true, $b$ is false, and evaluate each side of the "proposed equivalence.)

So clearly, $\lnot$ does not distribute over the parentheses: $$\lnot (\lnot b, a) \not \equiv (b, \lnot a)$$

This shouldn't be surprising, since $\lnot$ does not distribute over $a \rightarrow b$ to make $\lnot(a \rightarrow b)$ equivalent to $\lnot a \rightarrow \lnot b$. In your "connective" it follows that $\lnot(b, a) \not \equiv (\lnot b, \lnot a).$ Rather, $$\begin{align} \lnot (b,a) & \equiv \lnot (a \rightarrow b) \\ \\ &\equiv \lnot (\lnot a \lor b) \\ \\ & \equiv \lnot ( b \lor \lnot a)\\ \\ & \equiv \lnot (\lnot b \rightarrow \lnot a) \\ \\ &\equiv \lnot (\lnot a, \lnot b)\end{align}$$

Indeed, $(a, b) \equiv (\lnot b, \lnot a)$ by contraposition.

can Not logic move into this custom logic

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