Basic question on logic

Question

I have a slight problem in solving the following question. Let $P$ and $Q$ be statements. Which of the following strategies is "NOT" a valid way to show that "$P$ implies $Q$"?

1. Assume that $P$ is true, and then use this to show that $Q$ is true.
2. Assume that $Q$ is false, and then use this to show that $P$ is false.
3. Show that either $P$ is false, or $Q$ is true, or both.
4. Assume that $P$ is true, and $Q$ is false, and deduce a contradiction.
5. Assume that $P$ is false, and $Q$ is true, and deduce a contradiction.
6. Show that $P$ implies some intermediate statement $R$, and then show that $R$ implies $Q$.
7. Show that some intermediate statement $R$ implies $Q$, and then show that $P$ implies $R$.

I know that 5. is not a valid way but i'm really struggling with parts 6. and 7.

For part 6. I tried doing it this way:

Let $P$ be the statement "germany borders china", let $R$ be the statement "$2+2=4$" and $Q$ be the statement "pigs fly". Then $P$ will vacuously imply the intermediate statement $R$ but $R$ will not imply $Q$ because $R$ is true and $Q$ is false. Hence 6. is not a valid way.

Is this a correct way to check the validity of part 6. and 7.? If not, then what is?

Answer 1

It is no the same question but there is a partial answer to your question here. A simple way to see the tautologies is using truth tables. For instance, you can "prove" 6. by the truth table of $[(P \implies R) \land (R \implies Q)] \implies (P \implies Q)$.

The point 7. other way to say the same that 6., because you can write this by $[(R \implies Q) \land (P \implies R)] \implies (P \implies Q)$, since the conjunction is commutative.

Edit. If you want a "formal proof" for 6., you can try this (but I do not think is what you are looking for).

Using modus ponens (MP), i.e., if we have $P$ and $P \implies Q$, then we conclude $Q$. Now, we want to prove $P \implies Q$. Also, we know that $P \implies R$ and $R \implies Q$. Then

$$ \begin{array}{lll} 1 & \quad P \implies R & \quad\text{Assumption}\\ 2 & \quad R \implies Q & \quad\text{Assumption}\\ \quad 3 & \quad P & \quad\text{Hypothesis}\\ \quad 4 & \quad R & \quad\text{MP 1, 3}\\ \quad 5 & \quad Q & \quad\text{MP 2, 4}\\ 6 & \quad P \implies Q \end{array} $$

Answer 2

That is not a correct way to check the validity of 6. and 7.

You did pick a statement $R$ for 6. that did not imply $Q$, so that is not an application of the rule described in 6. As the comments mentioned, 6. works because of transitivity.

If you already showed that everything excluding 5. - 7. is a valid strategy to prove stuff, you can use 1. - 4. to show 6.

You can use 3. - "$P$ implies $Q$" is the same as "$P$ is false or $Q$ is true or both", and use that for "$P$ implies $R$" and "$R$ implies $Q$".

Another option is a proof via contradiction, e.g. option 4.:
Assume $P$ implies $R$, $R$ implies $Q$, but assume also that $P$ does not imply $Q$. What can you deduce about the truth values of $Q$ if $P$ is true or false?

Answer 3

(1) Answer

"Which of the following strategies is "NOT" a valid way to show that "P implies Q"?"

(5) is not a valid way to show that "P implies Q"

All the other alternatives are valid ways of showing that "P implies Q".

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(2) Justification

I first apologize if my symbolism may confuse to you, but I do assure you that they are necessary for a good understanding of your question.

Recall the definition of the meaning of the logical connective '$\to$':

$v(ϕ → ψ) = 1 ⇔ v(ϕ) \leq v(ψ)$

where $v$ is a valuation function.

Then it follows that (1), (2), (3), (4) are equivalent ways of stating that $v(ϕ → ψ) = 1$:

(1) $v(P → Q) = 1 ⇔ v(P) \leq v(Q)$ $ ⇔$ $ $if $v(P)=1$ then $v(Q)=1$

(2) $v(P → Q) = 1 ⇔ v(P) \leq v(Q)$ $ ⇔ $ if $v(Q)=0$ then $v(P)=0$

(3) $v(P → Q) = 1 ⇔ v(P) \leq v(Q)$ $ ⇔ $ $v(P)=0$ or $v(Q)=1$

(4) $v(P → Q) = 1 ⇔ v(P) \leq v(Q)$ $ ⇔$ $ v(P)=1$ and $v(Q)=0$ is a contradiction

Actually (6) and (7) too:

(6) $v(P → R) = 1$ and $v(R → Q) = 1$ $ ⇔$ $ v(P) \leq v(R) \leq v(Q)$ ⇔ v(P → Q) = 1$

Explanation: First, it says that we should assume that $P \to R$ is true, i.e. $v(P → Q) = 1$. Now this is the same as $v(P) \leq v(Q)$. Then since we assume that $R \to Q$ is also true then $v(P) \leq v(R) \leq v(Q)$. This shows that $P \to Q$ is true.

(7) goes simillarly:

(7) $v(R → Q) = 1$ and $v(P → R) = 1 ⇔ v(P) \leq v(R) \leq v(Q)$ ⇔ v(P → Q) = 1$

Explanation: it says we should assume that $R \to Q$ is true, i.e. $v(R → Q) = 1$. Now this is the same as $v(R) \leq v(Q)$. Then since we assume that $P \to R$ is also true then $v(P) \leq v(R) \leq v(Q)$. This shows that $P \to Q$ is true.

Answer 4

(5) is NOT valid. All other options are fine. $P \implies Q $ is FALSE iff $P$ is FALSE and $Q$ is TRUE.