for A → B condition to be invalid we have to show that B is false when A is true.
So we assume that A ( left side of → ) is true, then we note the value of B (right side of → ) if it is false then A → B is not a tautology .
a) ((a → b) ^ (b → c)) → (a → c)
left side of '→' is ((a → b) ^ (b → c)) which we assume to be true . for this to be true both (a → b) and (b → c) must be true. now if 'a' is false then right side (a → c) is true so there is need to check bcoz we have to show right side as false in case of fallacy
let 'a' is true then 'b' and 'c' must be true for left side ((a → b) ^ (b → c)) to be true .now as 'a' and 'c' is true (a → c) is true.
so it is a tautology.
c ) (a^b^c)→(c v a))
for left side (a^b^c) to be true 'a' ,'b' and 'c' must be true so right side (c v a) is also true.so it is a tautology.
d) a→(b→a)
for left side to be true 'a' must be true then (b→a) is also true bcoz we assume 'a' to be true. so it is a tautology.
b) (a ↔ c) →(~b→(a^c))
for left side to be true both ' a' and ' c' either be true or false.
In case when both are 0 then (a^c) is 0.so for b=0, a=0 and c=0 right side is false even if left side is true.As there is a case when 1→0 so it is not a tautology