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1) Show that ~P⇒ (QR) and Q⇒ (P v R) are logically equivalent using logical dentities. ~P⇒ (Q⇒ R) = Q = (PVR) 2) Show that ~(PAQ) ^ (~P v Q) and ~P are logically equivalent using logical identities. ~(PAQ) ^ (~P VQ) = ~P

1) Show that ~P⇒ (QR) and Q⇒ (P v R) are logically equivalent using logical dentities. ~P⇒ (Q⇒ R) = Q = (PVR) 2) Show that ~(PAQ) ^ (~P v Q) and ~P are logically equivalent using logical identities. ~(PAQ) ^ (~P VQ) = ~P

Advanced Engineering Mathematics
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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1) Show that ~P⇒ (QR) and Q (P v R) are logically equivalent using logical identities. ~P⇒ (QR) = Q = (PVR) 2) Show that ~(PAQ) ^ (~P v Q) and ~P are logically equivalent using logical identities. • (PAQ) ^ (~P v Q) = ~P
Transcribed Image Text:1) Show that ~P⇒ (QR) and Q (P v R) are logically equivalent using logical identities. ~P⇒ (QR) = Q = (PVR) 2) Show that ~(PAQ) ^ (~P v Q) and ~P are logically equivalent using logical identities. • (PAQ) ^ (~P v Q) = ~P
LOGICAL IDENTITIES Idempotence Commutative Associative De Morgan's Distributive Material Equivalence Involution Material Implication Exportation Identities P=PvP P= P&P (PvQ) = (Q vP) (P&Q)=(Q&P) (Pv (QR)) = ((PvQ) v R) (PA (QAR)) = ((PAQ) AR) (PvQ)=(PA-Q) (P&Q)=(-Pv -Q) (P&(Qv R)) = ((P&Q) v (P&R)) (Pv (Q&R)) = ((PvQ) ^ (PVR)) (P+Q) = ((P+Q) ^ (Q+P)) P-P (P+Q)=(PvQ) ((P&Q) +R)=(P+(Q+R)) (Pv TRUE)= TRUE (Pv FALSE) = P (P& TRUE) = P (P& FALSE)= FALSE (PA-P)= FALSE
Transcribed Image Text:LOGICAL IDENTITIES Idempotence Commutative Associative De Morgan's Distributive Material Equivalence Involution Material Implication Exportation Identities P=PvP P= P&P (PvQ) = (Q vP) (P&Q)=(Q&P) (Pv (QR)) = ((PvQ) v R) (PA (QAR)) = ((PAQ) AR) (PvQ)=(PA-Q) (P&Q)=(-Pv -Q) (P&(Qv R)) = ((P&Q) v (P&R)) (Pv (Q&R)) = ((PvQ) ^ (PVR)) (P+Q) = ((P+Q) ^ (Q+P)) P-P (P+Q)=(PvQ) ((P&Q) +R)=(P+(Q+R)) (Pv TRUE)= TRUE (Pv FALSE) = P (P& TRUE) = P (P& FALSE)= FALSE (PA-P)= FALSE
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