My first choice would be a truth table but if you must use "logical equivalences" you might start with "A implies B" is equivalent to "not B implies not A". Here that would be "not q implies not (p and (p implies q))"
Hi all,
I am currently working on showing that [p AND (p IMPLIES q)] IMPLIES q is a tautology and, really just don't know where to begin. Not expecting any answers, but to be pointed in the correct direction.
My first choice would be a truth table but if you must use "logical equivalences" you might start with "A implies B" is equivalent to "not B implies not A". Here that would be "not q implies not (p and (p implies q))"
Hello, michaelgg13!
Darn! Latex isn't working . . .
Show that: .[p ∧ (p → q)] → q
I'll list the steps.
I'll let you supply the reasons.
We have:
[p ∧ (~p ∨ q)] → q
[(p ∧ ~p) ∨ (p ∧ q)] → q
[F ∨ (p ∧ q)] → q
(p ∧ q) → q
~(p ∧ q) ∨ q
(~p ∨ ~q) ∨ q
~p ∨ (~q ∨ q)
~p ∨ T
. . .T