boolean-algebradiscrete mathematicslogicpropositional-calculus
I came across this problem, it asks to use logical equivalences (see image), show that $(p → r) ∨ (q → r)$ logically equivalent to the statement $(p ∧ q) → r$ (aka definition of biconditional)
After apply $p→q ≡ ¬p∨q$, I can't go any further can anyone show me any steps further?
Thank you
i think what you did until here:
[(Q∨P)∧T]∨¬P
is correct. But then you can simply forget about the T; the formula above is logically equivalent to
(Q∨P)∨¬P
which is simply
Q∨P∨¬P
Q∨T
T.
I managed to work out a solution, and I thought I would share even though it is rather hideous (picture posted to make formatting much more pleasant to read). 
Best Answer
$(p → r) ∨ ( q → r)=(¬p ∨ r) ∨ (¬q ∨ r)=¬p ∨ r ∨ ¬q ∨ r=(¬p ∨ ¬q) ∨ r=¬(p∧q)∨r=(p∧q) → r$