Prove that [r (q (~r →~p))] ≡ r (p∨ ~q) by using a series of logical equivalences

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p q r ~p ~q ~r ~r→~p (≡~p) q∧(~r→~p) (≡q∧~p) r∨(q∧(~r→~p))
(≡r∨(q∧~p))
p∨~q (≡~q∨p) r∧(p∨~q) (≡r∧(~q∨p)) 
0 0 0 1 1 1 1 0 0 1 0
0 0 1 1 1 0 1 0 1 1 1
0 1 0 1 0 1 1 1 1 0 0
0 1 1 1 0 0 1 1 1 0 0
1 0 0 0 1 1 0 0 0 1 0
1 0 1 0 1 0 0 0 1 1 1
1 1 0 0 0 1 0 0 0 1 0
1 1 1 0 0 0 0 0 1 1 1

0=false, 1=true.

This truth table covers all Boolean eventualities, but the two green columns (candidates for equivalence) do not match when (p,q,r)=(0,1,0) or (0,1,1). So the proposed equivalence is false.

by Top Rated User (1.2m points)

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