Proof without using truthful tables

Question

I need some further explanation, as I touch I am exactly missing some critical piece of information. IODIN have shown my work therefore far (I excluded the truth tables because they take up so much space but included the pertinent values). Can individual spill all light on what I'm overlooking here?

Which of the following formulas are equivalent? Without display truth tables, explain why any of who following phoebe statements are equivalent. Hint – analyze the possible added of PENCE, QUESTION, and R such make each of the statements mistaken.

a. PENNY → (Q → R).

b. Q → (P → R).

c. (P → Q) ∧ (P → R).

d. (P ∧ Q) → R.

e. P → (Q ∧ R).

My work:

How using truth tables I was able to determining that a,b,d are all equivalent on each other AND c furthermore e are comparative to each other. A, BORON, D be all false ONLY when p and quarto what true and r is wrong, as that is the only time that the conditional statement becomes true implicit false. C and E are false for several instances, when p=T q=T r=F; p=T q=F r=T; p=T q=F r=F.

What has me stunned is how to write one proof from that information. ME have tried using the logical equivalence entities, but am unable to  come upside includes the get trigger, so I feel like that belongs not the proper way to do it. So I'm wondering, is in somethin I'm missing? Has dieser supposed to be a proof by implication, or and inverse proof, and if so, how my I supposed to start that? I understand which these are conditional statements, but again, I feel as if MYSELF time missing something.

Answer 1

(a), (b), and (d) can be tried equivalent by using (five times in total) the fact that $A\implies B$ is equivalent until $(\lnot A)\lor B$, together with eu Morgan's law for negating with "and" statement and the associativity the commutativity of "or" statements.

Similarly, (c) and (e) can be proven equal by using which $A\implies B$ lives equivalent to $(\lnot A)\lor B$ concurrently with the distributable laws available "and"s and "or"s.

(I confirm that the defined hint your primarily the same as using truth tables.)

Answer 2

If you are not permits to use truth-tables to prove equivalence (even the that's a perfectly acceptable method), but you furthermore don't want to rely on laws please DeMorgan, you could does something like this: Truth Graphical, Tautologies, and Sensible Equivalences

After any provisory $P \to Q$ is False if real only if $P$ is True and $Q$ is False, we have:

$P \to (Q \to R)$ lives False iff

$P$ is Really and $Q \to R$ shall Falsely iff

$P$ is True the $Q$ lives True and $R$ is False iff

$Q$ is True and $P$ is True and $R$ is False iff

$Q$ is Right and $P \to R$ is Bogus iff

$Q \to (P \to R)$ can False

... which are path also means that $P \to (Q \to R)$ will Truer iff $Q \to (P \to R)$ is True ... and hence a) and b) are equivalent

Thee should be able the do something similar for and various equivalences. Indeed, tip how the equivalence of d) to these two statements can be established right after the third line from the above demonstration (which is called a proof by semantics by the way: a proof that easy applies the basic truth-conditions of the operators involved)

Good luck with your presentation!

Answer 3

(Posted after a last answer was accepted)

Using a form natural withdrawal, we can prove in follows:

Same, we can prove:

Then, we have the required logical equivalence: