Reason ер Vx(P(x)v Q(x)) P(a)vQ(a) P(a)vQ(a)v R(a) Premise Vx(-P(x)^Q(x))→R(x)) (-P(a)^Q(a))→R(a) -(-P(a)^Q(a))v R(a) P(a)v¬Q(a)v R(a) Premise V P(a)v R(a) -(-R(a))v P(a) - -R(a)→ P(a) . Vx(¬R(x)→P(x)
Reason ер Vx(P(x)v Q(x)) P(a)vQ(a) P(a)vQ(a)v R(a) Premise Vx(-P(x)^Q(x))→R(x)) (-P(a)^Q(a))→R(a) -(-P(a)^Q(a))v R(a) P(a)v¬Q(a)v R(a) Premise V P(a)v R(a) -(-R(a))v P(a) - -R(a)→ P(a) . Vx(¬R(x)→P(x)
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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Transcribed Image Text:5. [1.6
In the following argument, give a justification for each step.
Conclusion: Vx(-¬R(x)→P(x))
Vx(P(x)vQ(x))
Vx(-P(x)^Q(x))→ R(x))
Premises:
Step
1. Vx(P(x)vQ(x))
2. P(a)v Q(a)
3. Р(а)v@ (a)v R(a)
Reason
Premise
4. Vx((¬P(x)^Q(x))→ R(x))
5. (-P(a)^Q(a))→ R(a)
6. –(¬P(a)^Q(a) v R(a)
7. P(a)v¬Q(a)v R(a)
8. P(a)v R(a)
Premise
9. –(¬R(a)) v P(a)
10. —R(a) > Р(а)
11. Vx(¬R(x)→ P(x))
->
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