The solution they have given is lengthy. I think there must be some shortcut but unable to recall it. Please help. Thanks.
S2 hint: Apply the same trick as above but this time take R=T as in RHS we have ~R
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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