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
