Vp2Vp3) →q and apply a sequence of logical equivalences to obtain [(p₁ →q)^(p2 →q)^(p3 →q)]. !) Let r = p2 V p3, and show that (p1 V P2 V p3) →q is equivalent to (p₁ V r) → q. Note we really should only replace parenthesized subexpressions, so you should probably use the associativity of V to make (p2 V P3) appear as a subexpression first. 1) Apply the n = 2 version of the equivalence to your new expression from (i). You should obtain a compound expression with r→q as a subexpression. 1) Replace r with (p2 V p3), and apply the n = 2 equivalence again, this time to (p2 V P3) → q. ) Use associativity of A to show that your expression form (iii) is equivalent to the required form of [(P1 →q) ^ (P2 →q) ^ (P3 → q)].
Vp2Vp3) →q and apply a sequence of logical equivalences to obtain [(p₁ →q)^(p2 →q)^(p3 →q)]. !) Let r = p2 V p3, and show that (p1 V P2 V p3) →q is equivalent to (p₁ V r) → q. Note we really should only replace parenthesized subexpressions, so you should probably use the associativity of V to make (p2 V P3) appear as a subexpression first. 1) Apply the n = 2 version of the equivalence to your new expression from (i). You should obtain a compound expression with r→q as a subexpression. 1) Replace r with (p2 V p3), and apply the n = 2 equivalence again, this time to (p2 V P3) → q. ) Use associativity of A to show that your expression form (iii) is equivalent to the required form of [(P1 →q) ^ (P2 →q) ^ (P3 → q)].
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
discrete math
![Question B3:
To justify case analysis, the text notes that the following expression is a tautology (section 1.8.2):
[(P1 V P2 V ... V Pn) →q] → [(P₁ →q) ^ (P2 →q) ^ ^ (Pn →q)].
This is an infinite number of claims, asserting a different logical equivalence for each positive integer n.
(P1 V P2)→q¬(P₁ VP2) V q
= (P₁ ^ p2) V q
= q V (P₁ ^ P2)
= (qVp₁) ^ (q V
p2)
= (P₁ Vq) ^ ( P2 V q)
= (P₁ →q) ^ (P2 → q)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7355d962-0fce-4254-b762-05ebe0225361%2Fa780c90a-f412-49cd-940d-f5be94840000%2Fuu5t3la_processed.png&w=3840&q=75)
Transcribed Image Text:Question B3:
To justify case analysis, the text notes that the following expression is a tautology (section 1.8.2):
[(P1 V P2 V ... V Pn) →q] → [(P₁ →q) ^ (P2 →q) ^ ^ (Pn →q)].
This is an infinite number of claims, asserting a different logical equivalence for each positive integer n.
(P1 V P2)→q¬(P₁ VP2) V q
= (P₁ ^ p2) V q
= q V (P₁ ^ P2)
= (qVp₁) ^ (q V
p2)
= (P₁ Vq) ^ ( P2 V q)
= (P₁ →q) ^ (P2 → q)
![(b) To verify the n = 3 equivalence (without building a 16 row truth table), we need to start with
(p1 Vp2Vp3) →q and apply a sequence of logical equivalences to obtain [(p₁ →q)^(p2 →q)^(p3 →q)].
(i) Let r=p2 V p3, and show that (p₁ V p2 V P3) →q is equivalent to (p₁ Vr) →q.
Note we really should only replace parenthesized subexpressions, so you should probably use the
associativity of V to make (p2 V P3) appear as a subexpression first.
(ii) Apply the n = 2 version of the equivalence to your new expression from (i).
You should obtain a compound expression with r→q as a subexpression.
(iii) Replace r with (p2 V P3), and apply the n = 2 equivalence again, this time to (p2 V P3) →q.
(iv) Use associativity of A to show that your expression form (iii) is equivalent to the required form
of [(P1 →q) ^ (P2 →q) ^ (P3 → q)].
(c) How could you extend the approach in (b) to verify the n = 4 version of the tautology? What if you
needed a case analysis involving 5 or 742 cases?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7355d962-0fce-4254-b762-05ebe0225361%2Fa780c90a-f412-49cd-940d-f5be94840000%2Fhn70qys_processed.png&w=3840&q=75)
Transcribed Image Text:(b) To verify the n = 3 equivalence (without building a 16 row truth table), we need to start with
(p1 Vp2Vp3) →q and apply a sequence of logical equivalences to obtain [(p₁ →q)^(p2 →q)^(p3 →q)].
(i) Let r=p2 V p3, and show that (p₁ V p2 V P3) →q is equivalent to (p₁ Vr) →q.
Note we really should only replace parenthesized subexpressions, so you should probably use the
associativity of V to make (p2 V P3) appear as a subexpression first.
(ii) Apply the n = 2 version of the equivalence to your new expression from (i).
You should obtain a compound expression with r→q as a subexpression.
(iii) Replace r with (p2 V P3), and apply the n = 2 equivalence again, this time to (p2 V P3) →q.
(iv) Use associativity of A to show that your expression form (iii) is equivalent to the required form
of [(P1 →q) ^ (P2 →q) ^ (P3 → q)].
(c) How could you extend the approach in (b) to verify the n = 4 version of the tautology? What if you
needed a case analysis involving 5 or 742 cases?
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