Answer :
To complete the Law of Syllogism, we need to form a correct logical argument that connects the given statements properly. The Law of Syllogism states that if [tex]\( p \rightarrow q \)[/tex] and [tex]\( q \rightarrow r \)[/tex] are true statements, then [tex]\( p \rightarrow r \)[/tex] is also a true statement.
We are given:
- [tex]\( p \rightarrow q \)[/tex] as the first true statement
- We need to find what conditional [tex]\( \_\_\_\_\_\_\_\_\_\_\_\_ \)[/tex] will complete it
Given the options:
- [tex]\( q \rightarrow r \)[/tex]
- [tex]\( p \rightarrow 1 \)[/tex]
- [tex]\( q \rightarrow p \)[/tex]
- [tex]\( r \rightarrow q \)[/tex]
Let's analyze each option:
1. [tex]\( q \rightarrow r \)[/tex]: This says if [tex]\( q \)[/tex] is true, then [tex]\( r \)[/tex] is true. If we have both [tex]\( p \rightarrow q \)[/tex] and [tex]\( q \rightarrow r \)[/tex] as given and true statements, we can apply the Law of Syllogism to state [tex]\( p \rightarrow r \)[/tex].
2. [tex]\( p \rightarrow 1 \)[/tex]: This statement is not logically connected to [tex]\( p \rightarrow q \)[/tex] or would help in forming a syllogism.
3. [tex]\( q \rightarrow p \)[/tex]: This would be the converse of [tex]\( p \rightarrow q \)[/tex] and does not help in the application of the Law of Syllogism.
4. [tex]\( r \rightarrow q \)[/tex]: This would be the converse of the required conditional statement [tex]\( q \rightarrow r \)[/tex], again, not helping in the Law of Syllogism.
Therefore, the correct conditional to complete the Law of Syllogism is:
[tex]\[ q \rightarrow r \][/tex]
So, the correct option is:
[tex]\[ q \rightarrow r \][/tex]
We are given:
- [tex]\( p \rightarrow q \)[/tex] as the first true statement
- We need to find what conditional [tex]\( \_\_\_\_\_\_\_\_\_\_\_\_ \)[/tex] will complete it
Given the options:
- [tex]\( q \rightarrow r \)[/tex]
- [tex]\( p \rightarrow 1 \)[/tex]
- [tex]\( q \rightarrow p \)[/tex]
- [tex]\( r \rightarrow q \)[/tex]
Let's analyze each option:
1. [tex]\( q \rightarrow r \)[/tex]: This says if [tex]\( q \)[/tex] is true, then [tex]\( r \)[/tex] is true. If we have both [tex]\( p \rightarrow q \)[/tex] and [tex]\( q \rightarrow r \)[/tex] as given and true statements, we can apply the Law of Syllogism to state [tex]\( p \rightarrow r \)[/tex].
2. [tex]\( p \rightarrow 1 \)[/tex]: This statement is not logically connected to [tex]\( p \rightarrow q \)[/tex] or would help in forming a syllogism.
3. [tex]\( q \rightarrow p \)[/tex]: This would be the converse of [tex]\( p \rightarrow q \)[/tex] and does not help in the application of the Law of Syllogism.
4. [tex]\( r \rightarrow q \)[/tex]: This would be the converse of the required conditional statement [tex]\( q \rightarrow r \)[/tex], again, not helping in the Law of Syllogism.
Therefore, the correct conditional to complete the Law of Syllogism is:
[tex]\[ q \rightarrow r \][/tex]
So, the correct option is:
[tex]\[ q \rightarrow r \][/tex]
