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The solution they have given is lengthy. I think there must be some shortcut but unable to recall it. Please help. Thanks.

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1.  S1 hint: A→ B is only false when T→F case exist. So take B=False, and somehow try to make the LHS true. If u can make the LHS true then definitely it is not VALID. If u cant make the LHS true then the proposition is VALID. This is the shortcut, without having to solve the entire proposition. As u have only 3 variables A,B,C and B is already fixed to be F so there exists only 4 combinations for A and C which u can put in the LHS and check whether u are able to obtain T for LHS.

S2 hint: Apply the same trick as above but this time take R=T as in RHS we have ~R

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I was reading somewhere that if T->F assignment is not possible, then it is valid/tautology. But if such assignment is  possible, then it can be either contradiction or contingency. Is it true?
Also how to solve Question 7
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But if such assignment is  possible, then it can be either contradiction or contingency

This is absolutely true. Contradiction happens only when all the cases are false, but by T→ F we are only checking one of many cases, it may be the case that T→ T assignment is also possible for the same. Hence if T→F exists it is definitely not valid(i.e. NOT TAUTOLOGY) but it can be contradiction or contingency.

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I do not know how to solve 7. Never seen this type in GATE