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Q: Use logical equivalences to show that (p → q) → (notq → notp) is a tautology. Could you please show me this without just stating C2 rule (ϕ → ψ ≡ notψ → notϕ)   this is what we did in lecture but I am confused.  (p → q) → (notq → notp) (p -> q) -> (p->q) not(p->q) v (p->q) (p->q) v not(p->q) T

Q: Use logical equivalences to show that (p → q) → (notq → notp) is a tautology. Could you please show me this without just stating C2 rule (ϕ → ψ ≡ notψ → notϕ)   this is what we did in lecture but I am confused.  (p → q) → (notq → notp) (p -> q) -> (p->q) not(p->q) v (p->q) (p->q) v not(p->q) T

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Q: Use logical equivalences to show that (p → q) → (notq → notp) is a tautology.

Could you please show me this without just stating C2 rule (ϕ → ψ ≡ notψ → notϕ)

 

this is what we did in lecture but I am confused. 

(p → q) → (notq → notp)

(p -> q) -> (p->q)

not(p->q) v (p->q)

(p->q) v not(p->q)

T 

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