Show that ¬(¬p) and p are logically equivalent First, let's see a wordy interchanges the two truth values, so negating a second time interchanges them back to Since (p ∧ q) ←→ ¬p ∨ ¬q is T in all cases, therefore (p ∧ q) ≡ ¬p ∨ ¬q
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Section 1.2, selected answers
Math 114 Discrete Mathematics
D Joyce, Spring 2018
2.Show that:(:p) andpare logically equivalent.
First, let's see a wordy explanation.
Is:(:p) !pa tautology? Does it come out true no
matter what truth valuephas? There are two cases. In one case,pisT, so:pisF, and:(:p) isT; and since :(:p) andphave the same truth value,:(:p) !pcomes outT. In the other case,pisF, so:pisT, and:(:p) is F; and since:(:p) andpagain have the same truth value, :(:p) !pcomes outTin this case, too. Thus, in both cases,:(:p) !pcomes outT. Thus,:(:p) !pa tautology. Therefore,:(:p) andpare logically equivalent. That was a wordy explanation. You probably gave a much shorter one that's just as good, something like this: Negating interchanges the two truth values, so negating a second time interchanges them back to their original truth values. Since ::phas the same truth value asp, therefore:(:p) !p is a tautology. Another thing you could do is present a truth table like this:p:p::p:(:p) !pTFTT TTFT Then, since the last column only containsTs, it's a tautol- ogy.6.Use a truth table to verify this De Morgan's law:
:(p^q) :p_ :q:pq:(p^q) ! :p_ :qTTF T T F F FTFT F T F T T
FTT F T T T F
FFT F T T T T
Since (p^q) ! :p_ :qisTin all cases, therefore
(p^q) :p_ :q. You could stop one step earlier by noticing that since the columns for:(p^q) and:p_ :qare identical, therefore they're logically equivalent.12.Show that each implication in Exercise 10 is a tautol-
ogy without using truth tables. For these, you can use the logical equivalences given in tables 6, 7, and 8. a)[:p^(p_q)]!q. The following is a list of logically equivalent expressions. Since the last is a tautology, so is therst. Each step uses one of the logical equivalences in oneof the tables to substitute one subexpression for a logically
equivalent subexpression. [:p^(p_q)]!q :[:p^(p_q)]_q (::p_ :(p_q))_q (p_(:p^ :q))_q ((p_ :p)^(p_ :q))_q (T^(p_ :q))_q (p_ :q)_q p_(:q_q) p_T T There are many other routes you could take to reduce the original expression toT. This was just one of them. The other parts of 10 are similar. Here's how 10c might be proved. p^(p!q)!q p^(:p_q)!q (p^ :p)_(p^q)!q