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Is this a Tautology, contradiction or contingent?

$(m \Leftrightarrow m) \Leftrightarrow (m \Rightarrow m)$

My answer is that It is a tautology. But what is yours?

Can someone please explain with a truth table? Thank you so much!!!!!

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What does $\square$ mean? –  Git Gud Oct 27 '13 at 12:06
    
Sorry didnt understand you –  user2923027 Oct 27 '13 at 12:21
    
Why are people in the answers assuming $\square$ means $\iff$? –  Git Gud Oct 27 '13 at 12:23
    
but no one used that? Im sorry i might be missing something .. –  user2923027 Oct 27 '13 at 12:41
    
The connective I see on my screen is $\square$. It shows the same in two different browsers. In the edit window it also shows $\square$. –  Git Gud Oct 27 '13 at 12:42
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4 Answers

up vote 4 down vote accepted
m  |  m <--> m | m --> m | (m <-> m) <--> (m --> m) 
T  |     T     |    T    |             T            
F  |     T     |    T    |             T   

Note that each of $m \rightarrow m$ and $m \leftrightarrow m$ is a tautology (always true, regardless of the truth value of $m$), and hence, $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m)$$

is necessarily a tautology, as well, which means the following equivalence necessarily holds: $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m) \equiv T$$

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Thank you so much. Is there a way we can prove this by any other Rules such that we can derive it to T or F? IT should only be T i think. –  user2923027 Oct 27 '13 at 14:07
    
Yes, it is true, always true (since it is a tautology, which we've shown above). The columns show only $T$ above. Perhaps you're confusing my column breaks $|$ with an I? The last column to the right shows that the proposition is true when $m$ is true, and the proposition is true when $m$ is false. So it doesn't matter whether $m$ is true or false: the proposition is true in either case. Hence $(m \leftrightarrow m) \leftrightarrow (m \rightarrow m) \equiv T$ –  amWhy Oct 27 '13 at 14:11
    
Yes! Makes sense. Thank you Kind person! There is another thing I wanted to ask you and that is - In order to prove this from LHS to RHS - That answer will also be T. Right? To do that we have to use Rules like Distributive, Commutative? But I dont understand how I will prove this theoretically. I understood the turth table and the concept, but in written how can we prove this is right. –  user2923027 Oct 27 '13 at 14:17
    
@amWhy: Nice discussion and answer +1 –  Amzoti Oct 28 '13 at 1:19
    
The column break didnt even come on my browser initially and had one 'I' next to T. It all makes sense. –  user2923027 Oct 28 '13 at 1:25
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It is tauntology.

p  q  |  p->q
T  T  |   T
T  F  |   F
F  T  |   T
F  F  |   T

You always have the same value, that is m $$ (m\implies m) $$ is always true when the inputs are same.

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You are right. This is tauntology. –  depecheSoul Oct 27 '13 at 12:18
2  
"tauntology" strikes me as the "study of taunting"! ;-) –  amWhy Oct 27 '13 at 14:16
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This is a tautology.

Hint: If $m=T$ (True) then $(m \Leftrightarrow m) \Leftrightarrow (m \Rightarrow m) = (T \Leftrightarrow T) \Leftrightarrow (T \Rightarrow T) = T \Leftrightarrow T = T$. What if $m = F$ (False)?

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In Polish notation we can prove this from the following axioms under the rule of detachment:

{C$\alpha$$\beta$, $\alpha$} $\vdash$ $\beta$.

"C" symbolizes the conditional, and "E" the bi-conditional or equivalence operator. The well-formed formula has Emm on the left, Cmm on the right, and reads EEmmCmm.

Axiom 1-CmCnm Recursive Letter Prefixing

Axiom 2-CCmCnoCCmnCmo Self-Distribution

Axiom 3-CCmnCCnmEmn

The notation $\alpha$/$\beta$ indicates that the lower case letter $\alpha$ gets substituted uniformly with the well-formed formula $\beta$. The numeral symbols refer to the well-formed formulas. * functions as a separator, and - serves as a sign of detachment. So, 3 m/n * C2-4 would mean that in well-formed formula 3 we will substitute all instances of "m" with n, it has the same form (it is "the same" well-formed formula) as the well-formed formula as a conditional with 2 as the antecedent, and 4 the consequent. We will thus detach well-formed formula 4.

  2 o/m * 4

4 CCmCnmCCmnCmm

  4 * C1-5

5 CCmnCmm

  5 n/Cnm * C1-6

6 Cmm

  3 n/m*C6-7

7 CCmmEmm

  7*C6-8

8 Emm

  1 m/Cmm, n/Emm * C6-9

9 CEmmCmm

  3 m/Emm, n/Cmm * C9-10

10 CCCmmEmmEEmmCmm

  10 * C7-11

11 EEmmCmm

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