To solve this problem, let's carefully examine the logical implications presented and determine the conclusion.
### Details of the Problem:
1. We have two logical statements:
- [tex]\(x \Rightarrow y\)[/tex]: If [tex]\(x\)[/tex] is true, then [tex]\(y\)[/tex] must be true.
- [tex]\(y \Rightarrow z\)[/tex]: If [tex]\(y\)[/tex] is true, then [tex]\(z\)[/tex] must be true.
### Logical Implication and Transitivity:
The property of logical implication allows us to use the transitive property, which states:
- If [tex]\(x \Rightarrow y\)[/tex] and [tex]\(y \Rightarrow z\)[/tex], then [tex]\(x \Rightarrow z\)[/tex].
### Step-by-Step Solution:
#### Step 1: Understanding [tex]\(x \Rightarrow y\)[/tex]
- If [tex]\(x\)[/tex] is true, [tex]\(y\)[/tex] must be true.
- This does not say what happens if [tex]\(x\)[/tex] is false; [tex]\(y\)[/tex] can be either true or false.
#### Step 2: Understanding [tex]\(y \Rightarrow z\)[/tex]
- If [tex]\(y\)[/tex] is true, [tex]\(z\)[/tex] must be true.
- This does not say what happens if [tex]\(y\)[/tex] is false; [tex]\(z\)[/tex] can be either true or false.
#### Step 3: Combining the Implications
- Given [tex]\(x \Rightarrow y\)[/tex] and [tex]\(y \Rightarrow z\)[/tex], we need to establish whether [tex]\(x \Rightarrow z\)[/tex] holds.
If [tex]\(x \Rightarrow y\)[/tex] and [tex]\(y \Rightarrow z\)[/tex], then whenever [tex]\(x\)[/tex] is true, [tex]\(y\)[/tex] must be true (from [tex]\(x \Rightarrow y\)[/tex]), and since [tex]\(y\)[/tex] is true, [tex]\(z\)[/tex] must also be true (from [tex]\(y \Rightarrow z\)[/tex]).
Thus, whenever [tex]\(x\)[/tex] is true, [tex]\(z\)[/tex] must also be true, which means [tex]\(x \Rightarrow z\)[/tex] is a valid conclusion.
### Conclusion:
Given [tex]\(x \Rightarrow y\)[/tex] and [tex]\(y \Rightarrow z\)[/tex], the statement [tex]\(x \Rightarrow z\)[/tex] holds true by the transitive property of logical implication. Therefore, the correct answer is:
A. [tex]\(x \Rightarrow z\)[/tex]
