Question

In: Advanced Math

Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.)...

Let p and q be propositions.

(i) Show (p →q) ≡ (p ∧ ¬q) →F

(ii.) Why does this equivalency allow us to use the proof by contradiction technique?

Solutions

Expert Solution

(i) p and q are two propositions.
Let us prove that p q and ( p ¬q   F ) are equivalent using a Truth Table.

p q ¬q p ¬q F p ¬q F p q

0 0 1 0 0 1 1
0 1 0 0 0 1 1
1 0 1 1 0 0 0
1 1 0 0 0 1 1

Since the corresponding truth values of p q and p ¬q F are the same, hence (p q) ≡ (p ¬q) F

(ii) The technique of Proof by Contradiction is an Indirect Proof technique. To prove an implication P Q,
we assume (on the contrary) that P and NOT Q are both true. Under this antecedent, we prove that the Consequent is a Contradiction F (that is, a statement F which is always false, in some given logical system). Since a Contradiction arises, this therefore implies that P Q is indeed true. Hence, the equivalency  P Q ( P ¬Q   F ) allows us to use the proof by contradiction technique.


Related Solutions

Let p and q be the propositions p:You drive over 65 miles per hour. q: You...
Let p and q be the propositions p:You drive over 65 miles per hour. q: You get a speeding ticket. Write these propositions using p and q and logical connectives. a) You do not drive over 65 miles per hour. b) You drive over 65 miles per hour, but you do not geta 1 speeding ticket. c) You will get a speeding ticket if you drive over 65 miles per hour. d) If you do not drive over 65 miles...
Let T∈ L(V), and let p ∈ P(F) be a polynomial. Show that if p(λ) is...
Let T∈ L(V), and let p ∈ P(F) be a polynomial. Show that if p(λ) is an eigenvalue of p(T), then λ is an eigenvalue of T. Under the additional assumption that V is a complex vector space, and conclude that {μ | λ an eigenvalue of p(T)} = {p(λ) | λan eigenvalue of T}.
FOR EAICH PAIR OF PROPOSITIONS P AND Q STATE WHETHER ON NOT p=q p=(s→(p ∧¬r)) ∧...
FOR EAICH PAIR OF PROPOSITIONS P AND Q STATE WHETHER ON NOT p=q p=(s→(p ∧¬r)) ∧ ((p→(r ∨ q)) ∧ s), Q=p ∨ t
Let q and p be natural numbers, and show that the metric on Rq+p is equivalent...
Let q and p be natural numbers, and show that the metric on Rq+p is equivalent to the metric it has if we identify Rq+p with Rq × Rp.
Two compound propositions p and q in propositional logic are logically equivalent if . . ..
Complete the following statements.Two compound propositions p and q in propositional logic are logically equivalent if . . ..An argument form in propositional logic is valid if . . ..A theorem is a statement that . . ..A statement that is assumed to be true is called a(n) . . ..A proof is a valid argument that . . ..
1. Show that the argument (a) p → q       q → p       therefore p...
1. Show that the argument (a) p → q       q → p       therefore p V q       is invalid using the truth table. ( 6 marks ) (b) p → q       P       therefore p        is invalid using the truth table. ( 6 marks ) (c) p → q       q → r        therefore p → r         is invalid using the truth table. ( 8 marks )
Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then...
Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then P = Q.
Let a < c < b, and let f be defined on [a,b]. Show that f...
Let a < c < b, and let f be defined on [a,b]. Show that f ∈ R[a,b] if and only if f ∈ R[a, c] and f ∈ R[c, b]. Moreover, Integral a,b f = integral a,c f + integral c,b f .
Show that, for any events E and F, P(E ∪ F) = P(E) + P(F) −...
Show that, for any events E and F, P(E ∪ F) = P(E) + P(F) − P(E ∩ F). Only use the probability axioms and indicate which axiom you use where
7. Let E be a finite extension of the field F of prime characteristic p. Show...
7. Let E be a finite extension of the field F of prime characteristic p. Show that the extension is separable if and only if E = F(Ep).
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT