Prove that the given proposition is logically equivalent to p. Apply appropriate laws of equivalences for proving. Provide handwritten solution following the table format provided. Upload a clear copy of the image, file should only be an image type or pdf. pV(qA (-p→q)) = p Resulting propositions Applied Law of Equivalence Use the following tables as references for the equivalences. TABLE 6 Logical Equivalences. Equivalence Name PAT=P Identity laws pVF= p TABLE 7 Logical Equivalences Involving Conditional Statements. TABLE 8 Logical Equivalences Involving Biconditionals. pVT=T Domination laws PAF=F Pvp=p Idempotent laws P q =-p Vq p + q = (p q)^ (q → p) PAp=p d- e b = b +d Pvq =-p→9 -(-p) =P Double negation law p+q = -p ++ pvq =q vp Commutative laws P + q = (p^q) v(-p^¬q) -(p + q) = p → - p^q =-(p → -) (p- q) = p^ng (p vq) vr=pv (q vr) (paq)ar=PA (qAr) Associative laws pv (q ar) = (p v q)A (p vr) PA(q vr) = (png) v (par) (p 4)^(p →r) = p (q ^r) (p r)^(q r) = (p v q) → r Distributive laws -(png) = -p v-g (p v q) = -p A-g (p 4)v (p r) = p (q vr) (p r)V (q → r) = (p ^q) → r De Morgan's laws pv (pnq) = p PA(pvq) =p Absorption laws pv-p=T Negation laws PA-p=F
Prove that the given proposition is logically equivalent to p. Apply appropriate laws of equivalences for proving. Provide handwritten solution following the table format provided. Upload a clear copy of the image, file should only be an image type or pdf. pV(qA (-p→q)) = p Resulting propositions Applied Law of Equivalence Use the following tables as references for the equivalences. TABLE 6 Logical Equivalences. Equivalence Name PAT=P Identity laws pVF= p TABLE 7 Logical Equivalences Involving Conditional Statements. TABLE 8 Logical Equivalences Involving Biconditionals. pVT=T Domination laws PAF=F Pvp=p Idempotent laws P q =-p Vq p + q = (p q)^ (q → p) PAp=p d- e b = b +d Pvq =-p→9 -(-p) =P Double negation law p+q = -p ++ pvq =q vp Commutative laws P + q = (p^q) v(-p^¬q) -(p + q) = p → - p^q =-(p → -) (p- q) = p^ng (p vq) vr=pv (q vr) (paq)ar=PA (qAr) Associative laws pv (q ar) = (p v q)A (p vr) PA(q vr) = (png) v (par) (p 4)^(p →r) = p (q ^r) (p r)^(q r) = (p v q) → r Distributive laws -(png) = -p v-g (p v q) = -p A-g (p 4)v (p r) = p (q vr) (p r)V (q → r) = (p ^q) → r De Morgan's laws pv (pnq) = p PA(pvq) =p Absorption laws pv-p=T Negation laws PA-p=F
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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