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p→q =¬p vq p→q =¬q →¬p pvq =-p→ q p^q=¬(p→¬g) |¬(p→q) = p^¬qg |(p→g)^(p→r)=p→(qnr) (p→r)^(q→r)=(pvq)→r (p→q)v(p→r)=p→(qvr) (p→r)v(q→r)=(p^q)→r

p→q =¬p vq p→q =¬q →¬p pvq =-p→ q p^q=¬(p→¬g) |¬(p→q) = p^¬qg |(p→g)^(p→r)=p→(qnr) (p→r)^(q→r)=(pvq)→r (p→q)v(p→r)=p→(qvr) (p→r)v(q→r)=(p^q)→r

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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TABLE 7 Logical Equivalences Involving Conditional Statements. For every of 9 statements prove its correctness.

The last 10th question: How many ways are there to select three unordered elements from a set with five elements when repetition is allowed? Find answer and give explanation.

More Logical Equivalences TABLE 7 Logical Equivalences Involving Conditional Statements. TABLE 8 Logical Equivalences Involving Biconditional Statements. p→q =-p vq p→q =¬q –→¬p p+q = (p→q)^(q → p) pvq =-p→ q p^q=-(p→-q) -(p→q)= p^¬g (p→g)^(p→r)= p→(qnr) (p→r)^(q→r) =(pvq)→r (p→q)v(p→r)= p→ (p→r)v(q→r)=(png)→r p+q = (p^g)v(-pn¬9) -(p+q) = p+¬q (qvr)
Transcribed Image Text:More Logical Equivalences TABLE 7 Logical Equivalences Involving Conditional Statements. TABLE 8 Logical Equivalences Involving Biconditional Statements. p→q =-p vq p→q =¬q –→¬p p+q = (p→q)^(q → p) pvq =-p→ q p^q=-(p→-q) -(p→q)= p^¬g (p→g)^(p→r)= p→(qnr) (p→r)^(q→r) =(pvq)→r (p→q)v(p→r)= p→ (p→r)v(q→r)=(png)→r p+q = (p^g)v(-pn¬9) -(p+q) = p+¬q (qvr)
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