Here is another approach...
problem 1 solution: $[(p\to q)\wedge (q\to r)]\to (p\to r)$
The expression is not a tautology if we find any 1 case when it is false.
As it's an implication, it is false IFF ( LHS = T and RHS = F )
Let's check if it's possible.
consider RHS. We have to make it False.
p-->r is FALSE only when ( p = T AND r = F )
So fix the values as p=T and r=F
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Now consider LHS. We have to make it TRUE
LHS is true only when both the terms of the ( ^ ) are TRUE
So check if it is possible.
(q-->r) is NOT TRUE IFF q=T and r=F
r = F from out assumption. So, q cannot be T (because we want the whole term to be TRUE) So, q=F
Thus far we have, p=T, r=F,q=F
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Now, consider the first term of ( ^ ).
p-->q should be TRUE (because we want to make the entire LHS TRUE)
This is possible IFF p is NOT TRUE WHILE q IS FALSE
but our assumption is p=T, r=F,q=F.
Therefore, p-->q is FALSE.
Which makes LHS False.
Hence, we cannot make RHS=F and LHS=T at the same time. Which means the given expression can never be false == > tautology.
problem 2 solution: $[(p \vee q) \wedge (p \to r) \wedge (q \to r)] \to r$
Again using the same approach,
This is an implication which can be False IFF ( LHS =T AND RHS = F )
Make RHS False. i.e. r = F
Now consider LHS. We have to make LHS =TRUE
Which is possible only when all the 3 terms in LHS are TRUE (as they are ANDed)
Now take q-->r is LHS.
r = F from our assumption. Therefore q cannot be T. Which means q is F.
So far we have r = F, q = F
Now take p-->r
again r=F. Therefore p cannot be T. Which means p is False.
So far we have, q=F, r=F, p=F
Now take p ∨ q
We already have p=F, q=F
Which means p ∨ q cannot be True.
Hence, the LHS != True
Which means, we can never make LHS=T and RHS=F
in the given expression. So it can never be False
==> it is a tautology