Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine which conditional completes the Law of Syllogism, let's first understand what the Law of Syllogism is:
The Law of Syllogism states that if we have two conditionals: [tex]\( p \rightarrow q \)[/tex] and [tex]\( q \rightarrow r \)[/tex], then we can infer a third conditional: [tex]\( p \rightarrow r \)[/tex].
Given:
- [tex]\( p \rightarrow q \)[/tex] (Read as: If [tex]\( p \)[/tex] then [tex]\( q \)[/tex])
We are supposed to identify the correct choice that should be in the form of [tex]\( q \rightarrow r \)[/tex] so that we can deduce [tex]\( p \rightarrow r \)[/tex].
Let's analyze each choice:
1. [tex]\( q \rightarrow 1 \)[/tex]
- This statement means if [tex]\( q \)[/tex] then [tex]\( 1 \)[/tex]. However, [tex]\( 1 \)[/tex] is not in our original statements [tex]\( p, q, r \)[/tex]. This does not help infer [tex]\( p \rightarrow r \)[/tex].
2. [tex]\( q \rightarrow p \)[/tex]
- This means if [tex]\( q \)[/tex] then [tex]\( p \)[/tex]. It does not match the required form [tex]\( q \rightarrow r \)[/tex] and does not help us infer [tex]\( p \rightarrow r \)[/tex].
3. [tex]\( r \rightarrow q \)[/tex]
- This means if [tex]\( r \)[/tex] then [tex]\( q \)[/tex]. This is actually the reverse of what we need. However, thinking of contrapositive logic, it is equivalent to saying [tex]\( \neg q \rightarrow \neg r \)[/tex] which is still not in our required form [tex]\( q \rightarrow r \)[/tex].
4. [tex]\( p \rightarrow 1 \)[/tex]
- This means if [tex]\( p \)[/tex] then [tex]\( 1 \)[/tex]. Similar to choice 1, [tex]\( 1 \)[/tex] is not in our original statements [tex]\( p, q, r \)[/tex]. Hence, this does not help infer [tex]\( p \rightarrow r \)[/tex].
Among these conditionals, analyzing [tex]\( r \rightarrow q \)[/tex] (choice 3), let's think step-by-step:
If we have:
- [tex]\( p \rightarrow q \)[/tex] (given)
- [tex]\( r \rightarrow q \)[/tex] (choice 3)
Using contrapositive of [tex]\( r \rightarrow q \)[/tex]:
- [tex]\( \neg q \rightarrow \neg r \)[/tex]
From original [tex]\( p \rightarrow q \)[/tex]:
- Contrapositive of [tex]\( p \rightarrow q \)[/tex] is [tex]\( \neg q \rightarrow \neg p \)[/tex]
If [tex]\( \neg q \rightarrow \neg p \)[/tex] holds true and we assert [tex]\( \neg q \rightarrow \neg r \)[/tex] holds, logically, we can deduce [tex]\( p \rightarrow r \)[/tex].
Therefore, conditional [tex]\( r \rightarrow q \)[/tex] is logically equivalent to helping in constructing [tex]\( p \rightarrow r \)[/tex].
Thus, the correct choice is [tex]\( \boxed{3} \)[/tex]: r \rightarrow q
The Law of Syllogism states that if we have two conditionals: [tex]\( p \rightarrow q \)[/tex] and [tex]\( q \rightarrow r \)[/tex], then we can infer a third conditional: [tex]\( p \rightarrow r \)[/tex].
Given:
- [tex]\( p \rightarrow q \)[/tex] (Read as: If [tex]\( p \)[/tex] then [tex]\( q \)[/tex])
We are supposed to identify the correct choice that should be in the form of [tex]\( q \rightarrow r \)[/tex] so that we can deduce [tex]\( p \rightarrow r \)[/tex].
Let's analyze each choice:
1. [tex]\( q \rightarrow 1 \)[/tex]
- This statement means if [tex]\( q \)[/tex] then [tex]\( 1 \)[/tex]. However, [tex]\( 1 \)[/tex] is not in our original statements [tex]\( p, q, r \)[/tex]. This does not help infer [tex]\( p \rightarrow r \)[/tex].
2. [tex]\( q \rightarrow p \)[/tex]
- This means if [tex]\( q \)[/tex] then [tex]\( p \)[/tex]. It does not match the required form [tex]\( q \rightarrow r \)[/tex] and does not help us infer [tex]\( p \rightarrow r \)[/tex].
3. [tex]\( r \rightarrow q \)[/tex]
- This means if [tex]\( r \)[/tex] then [tex]\( q \)[/tex]. This is actually the reverse of what we need. However, thinking of contrapositive logic, it is equivalent to saying [tex]\( \neg q \rightarrow \neg r \)[/tex] which is still not in our required form [tex]\( q \rightarrow r \)[/tex].
4. [tex]\( p \rightarrow 1 \)[/tex]
- This means if [tex]\( p \)[/tex] then [tex]\( 1 \)[/tex]. Similar to choice 1, [tex]\( 1 \)[/tex] is not in our original statements [tex]\( p, q, r \)[/tex]. Hence, this does not help infer [tex]\( p \rightarrow r \)[/tex].
Among these conditionals, analyzing [tex]\( r \rightarrow q \)[/tex] (choice 3), let's think step-by-step:
If we have:
- [tex]\( p \rightarrow q \)[/tex] (given)
- [tex]\( r \rightarrow q \)[/tex] (choice 3)
Using contrapositive of [tex]\( r \rightarrow q \)[/tex]:
- [tex]\( \neg q \rightarrow \neg r \)[/tex]
From original [tex]\( p \rightarrow q \)[/tex]:
- Contrapositive of [tex]\( p \rightarrow q \)[/tex] is [tex]\( \neg q \rightarrow \neg p \)[/tex]
If [tex]\( \neg q \rightarrow \neg p \)[/tex] holds true and we assert [tex]\( \neg q \rightarrow \neg r \)[/tex] holds, logically, we can deduce [tex]\( p \rightarrow r \)[/tex].
Therefore, conditional [tex]\( r \rightarrow q \)[/tex] is logically equivalent to helping in constructing [tex]\( p \rightarrow r \)[/tex].
Thus, the correct choice is [tex]\( \boxed{3} \)[/tex]: r \rightarrow q
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
