How to show that (p ⇒ q) ∧ (¬r ⇒ ¬q) equals (p ⇒ ¬r) ∨ (q ∧ ¬q)?
When I worked this out I found that they weren't equal. What I had was
P Q R (P => Q) ^ (¬R => ¬Q)
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
P Q R (P => ¬R) v (Q ^ ¬Q)
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
where 1 the same as T, for true, 0 is the same as F, for false. => is the symbol for implies, ¬ is the symbol for not, v is the symbol for 'or', and ^ is the symbol for 'and'
Update:Actually I just realized that we don't have to prove that they are equal, only that we have to prove whether the statement is true or not... so I can simply say its false lol...
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Verified answer
1. it is (p ⇒ r) and not : (p ⇒ ¬r)
2. (p ⇒ q) ∧ (¬r ⇒ ¬q) cannot be
"equal to" (means "equivalent to", with a <=>) : (p ⇒ r) ∨ (q ∧ ¬q)
there is only a : " => "
because : (q ∧ ¬q) is always False (=0) : q and ¬q cannot be True together
and if
[ (p ⇒ q) ∧ (¬r ⇒ ¬q) ] => [ (p ⇒ q) ∧ (q => r) ] (important: contraposal of (¬r ⇒ ¬q) )
and
[ (p ⇒ q) ∧ (q => r) ] => (p=>r) (transitivity)
finally
[ (p ⇒ q) ∧ (¬r ⇒ ¬q) ] => (p=>r)
so also
[ (p ⇒ q) ∧ (¬r ⇒ ¬q) ] => [(p=>r) V (q ∧ ¬q)]
(adding (q ∧ ¬q) is "not more expensive" because always False )
but on the reverse way:
from : (p=>r)
you cannot get : [ (p ⇒ q) ∧ (q => r) ]
hope it' ll help !!