Let a, b, c, d be propositions. Assume that the equivalences a ↔ (b V ~b) and b ↔ c hold. Then the truth value of the formula (a ∧ b) → ((a ∧ c) ∨ d) is always?
I solved using hint -
a ↔ (b V ~b)
a ↔ 1
And,
b ↔ c
Given equation,
(a ∧ b) → ((a ∧ c) ∨ d)
~(a ∧ b) ∨ ((a ∧ c) ∨ d)
On putting value of above two we have,
~(1 ∧ c) ∨ ((1 ∧ c) ∨ d)
(~c ∨ c )∨ d
1 ∨ d
1
But this answer is bit difficult. So I want to ask is there any other way to solve this.
There are always different ways to solve it. My solution is less formal than yours but nevertheless valid reasoning:
$a \leftrightarrow (b\lor \lnot b) \Rightarrow a $ is true.
Now there are two cases.
Assume, $b$ is false. The left hand side, $(a \land b)$ is then false and thus, the whole formula is true.
Assume, $b$ is true. Then $c$ is true as well (because of $b \leftrightarrow c$) and so the right hand side, $(a \land c) \lor d$ is true and thus, the whole formula is true.
Note that a formula of kind $F \rightarrow G$ is only false when $F$ is true and $G$ is false. Whenever F is false or whenever G is true, the formula is true.
I would say what you do is about as simply as you can get it! Yes: use Complement to substitute $1$ for $b \lor \neg b$, and hence substitute $1$ for $a$. Also: substitute $b$ for $c$ (or vice versa). And finally, rewrite the $\rightarrow$ into $\lor$ so you can use Complement again and get your answer ... but you did all that! I really see no simpler way to do this.
