Question

5.) Logic problems. List the truth values of the two statements of problems
(a) and the truth value of the statement of problem (b) in terms of the truth
values of P and Q.
a.) Determine if the following pairs of logical statements are equivalent.
Show why they are or are not equivalent.
i. (( P) ^ Q) ) P;
ii. (P _ ( Q)):
b.) Determine if the following statement is a tautology. Show why it is
or is not.
(P ) Q) _ (Q ) P)

5.) Logic problems. List the truth values of the two statements of problems (a) and the truth value of the statement of problem (b) in terms of the truth values of P and Q. a.) Determine if the following pairs of logical statements are equivalent. Show why they are or are not equivalent. i. ((~P) AQP; ii. ~ (PV-Q)). b.) Determine if the following statement is a tautology. Show why it is or is not. ( PQ) VQP)

Answer 1

In this problem I Have Given Both Approaches , By Truth Table And By Identities And Laws Both !

I Am Attaching One Truth Table Of P=>Q For Understanding !

Explanation :

Р Q PBQ T F F T T т F T T Fwith F T Scanned EmScanner

True when both P and a true Solution : & some Important Identities and -> Symbols: PAQ pand Q Logical cal rules PVQ P or Q True when por a True O P=>Q. If p, then d IFP true then ch also true NP not - P True when is false. not - Q True when ta is False Now To wiu truth values solve this problem we Make Truth Table for corresponding of statement. (a.) I ((~P) ^Q) => & lis) ~(PV [~Q) Entote p q 7 - up vq] up NQ (NP) AQ PVNQ) IN/PV(NQ) (cup hop a TIT T T T F F T T T TI T T T T T F T F T CamScanner

we From the truth table statement i not Equivalent can Conclude and statemen lii) that are Notei- Actually both statements are NOT of Each o logically other. 16. Truth table For. op (P => Q) V ( Q => P) P => Q QP (PQ) v (Q>>p) IT T T T T F F T T F T T T T T T T so Givens Expression is a Since AU Entries in the last coloum are (T). tautology, An Expression gines only then Statement is known cs asanne Tauto logy, True

Now and can we Solve both parts Some Laws, P using -> @ ([mP) A B) = ? (NP) A Q) v p pivoval v P = E NXVY " ~(XAQ ( XJ V (na De Morgans law (oro (rup) = p) =) [PVNQ)] Up PVNQ) v P - PUPVNO). PVNQ) xvy = yvx So, on Simplication gives 9 statement to it and statement- PVNQ) is 1~/PV(NQ)) so both are not Equivalent. CS Scanned with CamScanner

(P => Q) VIQ = D) ۷ مه ) ~Q) v NQ UP) NP va v NQ NP :: X Y E NXV y - : x vy уух NX VX = T (rue) (NP v pv (Quna) P T v I so given iven statement is is Aways tautology. CS Scanned with CamScanner
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