proposition logic problem

Question

below is proposition

"When A holds, if B holds C holds, if B does not hold C also does not holds. Regardless of B,C if A does not hold, then D does not hold. "

Write the above sentence with a propositional formula.

$A \implies (B \implies C)$

$A \implies ( \sim B \implies \sim C)$

$\sim A \implies \sim D$

Suppose that the sentence with proposition above is holds. At this time, determine the truth of "If both C and D hold, then B holds."

is this mean i have to find :

$(A \implies (B \implies C) \land (A \implies ( \sim B \implies \sim C)) \land (\sim A \implies \sim D)) \implies ((C \land D) \implies B)$

is tautology or not?

I draw the table , but there is false so it means the proposition does not holds?

Answer 1

$$C\implies B\equiv \neg C\lor B$$ $$D\implies(C\implies B)\equiv \neg D\lor\neg C\lor B\equiv \neg(C\land D)\lor B\equiv(C\land D)\implies B$$ We have two true sentences in your conjunction. $Q.E.D.$ Edited: This is the final step following my last statement in comments, but the core of the problem is here: Looking at $A$ I wrote has to hold necessarily (transitivity of the implication) @kili asked for an explanation when rewriting the implication using $\lor\;\&\;\land$. In that case, it appears, the conjunction has to be true (read more on CNF & DNF), it can be checked by plugging into all the truth values $1,0$ for each of the subsentences.

Answer 2

We have to check whether the premises entail the conclusion, i.e.

$A ⟹ (B ⟹ C), A ⟹ (\lnot B ⟹ \lnot C), \lnot A ⟹ \lnot D \vDash (C \land D) ⟹ B$

which is the same as checking whether:

$(A ⟹ (B ⟹ C)) \land (A ⟹ (\lnot B ⟹ \lnot C)) \land( \lnot A ⟹ \lnot D) ⟹ ((C \land D) ⟹ B)$

is a tautology.

We can avoid the full Truth Table development considering the following approach: assume not, i.e. assume that the conclusion is FALSE and the premises are all TRUE.

In order for the conclusion: $(C \land D) ⟹ B$ to be false, we must have a line in the TT where:

$C$ and $D$ are both TRUE and $B$ is FALSE.

What happens to the premises with this line of the TT ?

With $D$ TRUE we must have also $A$ TRUE, in order to satisfy the third premise: $\lnot A ⟹ \lnot D$.

Finally, what happens to 1st and 2nd premises with $A$ TRUE ?

Consider 1st one: $A ⟹ (B ⟹ C)$. We have that $A$ and $B$ are TRUE while $C$ is FALSE. This means: $\top ⟹ (\top ⟹ \bot)$ which is FALSE, and this contradicts our assumption that all premises are TRUE.

Conclusion: the assumption that the formula is not a tautology led us to a contradiction, and thus the formula must be a tautology.

Answer 3

You have :

(1) A --> ( B --> C)

(2) A --> ( ~B --> ~C)

(3) ~ A --> ~ D

The goal is : ( C & D) --> B

So, let's suppose (C & D) is true, and let's try to derive that B is true.

• If ( C& D) is true, then, in particular, D is true. But, by contraposition, (3) is equivalent to (D --> A) ( once one has applied double negation). From D and ( D -->A) one can derive that A is true ( by modus ponens).

• Now, since A is true, using (2) , it follows that ( ~B --> ~C) is true. Using contraposition ( and double negation) this is equivalent to ( C --> B) .

• But from our initial hypothesis ( namely: C&D) it follows that C is true.

• From (C -->B) and C, it follows that B is true, as desired.

• The conditional proof rule allows us to say that : (C&D) implies B.