Logic: Tautology

yoyobarn · **Joined:** Fri, 12 Nov 2010 07:00:12 UTC **Posts:** 48

Which of the following is NOT a tautology?

(A) $\lnot(\lnot{}A)\leftrightarrow{}A$

(B) $A\lor{}(\lnot{}A)$

(C) $[(A\land{}B)\rightarrow{}C]\leftrightarrow{}[A\rightarrow{}(B\lor{}C)]$

(D) $\lnot{}(C\land{}D)\leftrightarrow{}\lnot{}C\lor{}\lnot{}D$

(E) $A\lor{}(B\land{}C)\leftrightarrow(A\land{}B)\lor{}(A\land{}C)$

Source: Cracking the Mathematics GRE, Princeton Review..

I would think the answer is option C AND E?

C is wrong since
A: True
B: False
C: False
would produce a true value for the "left-hand side" while producing a false value for the 'right hand side'

E is also wrong since
A: True
B: False
C: False
would produce the necessary counterexample.

Is there anything wrong with my above reasoning?

I am very confused by the book's answer: Finally, choice (E) is the Distributive Property for sentential logic (clearly wrong??), so the statement is always true regardless of the values of sentences A, B, and C.

Soroban · **Posted:** Tue, 18 Jan 2011 16:59:53 UTC

Hello, yoyobarn!

I agree with you.

$(E)\;A \vee{(B \wedge C)\;\leftrightarrow\;(A \wedge B)\vee (A \wedge C)$ . is not the Distributive Property.

Probably a typo . . . a spectacular one!

Either the left side is: . $A \wedge (B \vee C)$

. .or the right side is: . $(A \vee B) \wedge (A \vee C)$

yoyobarn · **Joined:** Fri, 12 Nov 2010 07:00:12 UTC **Posts:** 48

Thanks a lot, Soroban!

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Logic: Tautology

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