22.214.171.124 Practicing equivalencies
Rewrite each of these statements five times using any rules of equivalence that can apply to them.
1. A > (B . ~ C)
2. H v ( L ≡ T)
3. D > [(G . H) > F]
4. ~[M v ~(N . O)] ≡ P
5. [ ~ (P v B) . (~H . ~ J)] > W
The biggest lesson to take away from this exercise is that rules of equivalence can be applied at any location, the main operator or any subordinate operator. So for instance, DN can be applied to #1 in several ways:
1. A > (B . ~ C)
~~[A > (B . ~C)] It can be applied to the whole compound statement.
~~A > (B . ~ C) It can be applied to any simple statement within the whole compound statement.
~~[~~A > ~~(~~B . ~ C)] It could be applied to every statement (to every simple statement and to every compound statement)! In practice, one would not do this, but the point here is to see the possibility, to see that any time one adds two tildes or deletes two tildes, one creates a statement equivalent to the original.
#1 can also be rewritten using CM:
A > (~C . B)
and both this commuted version and the original would allow being rewritten by TRAN (flipping the order of the antecedent and consequent, negating both):
~(~C . B) > ~ A
~(B . ~C) > ~ A
Finally IMP could rewrite the original (or either of the transposed versions) by changing the conditional to a negated disjunction
A > (B . ~ C) would become: ~A v (B . ~C)
or ~(~C . B) > ~ A would become: ~~(~C . B) v ~ A. And DN can remove those two tildes: (~C . B) v ~ A. Now the statement is a candidate for DS, instead of MP or MT.
This form (a disjunction with a conjunction as one of the disjuncts) can be rewritten by Distribution:
~A v (B . ~C) becomes (~A v B) . (~A v ~C) DIST
And the right hand conjunct of this (a disjunction of negatives) is an obvious candidate for De Morgan’s rule:
(~A v B) . ~(A . C) DM. Now we could apply SM or CM and SM