Answer :
To solve the problem, we will use logical reasoning based on the transitive property of implication.
The transitive property of implication states that if [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex] are both true, then [tex]\( x \Rightarrow z \)[/tex] must also be true.
Let's analyze the given statements one by one:
A. [tex]\(\neg x \Rightarrow \neg z\)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is false, then [tex]\( z \)[/tex] must also be false.
- This does not necessarily follow from the given premises [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex].
B. [tex]\( x \Rightarrow z \)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is true, then [tex]\( z \)[/tex] must also be true.
- This is directly supported by the transitive property of implications since [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex], hence [tex]\( x \Rightarrow z \)[/tex] must be true.
C. [tex]\( \neg x \Rightarrow z \)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is false, then [tex]\( z \)[/tex] must be true.
- There is no implication rule that directly supports this statement from [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex].
D. [tex]\( z \Rightarrow x \)[/tex]:
- The statement suggests that if [tex]\( z \)[/tex] is true, then [tex]\( x \)[/tex] must also be true.
- There is no implication rule that supports this reversal of the initial implications.
By considering the transitive property of implication, the only statement that must be true given [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex] is:
B. [tex]\( x \Rightarrow z \)[/tex].
The transitive property of implication states that if [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex] are both true, then [tex]\( x \Rightarrow z \)[/tex] must also be true.
Let's analyze the given statements one by one:
A. [tex]\(\neg x \Rightarrow \neg z\)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is false, then [tex]\( z \)[/tex] must also be false.
- This does not necessarily follow from the given premises [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex].
B. [tex]\( x \Rightarrow z \)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is true, then [tex]\( z \)[/tex] must also be true.
- This is directly supported by the transitive property of implications since [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex], hence [tex]\( x \Rightarrow z \)[/tex] must be true.
C. [tex]\( \neg x \Rightarrow z \)[/tex]:
- The statement suggests that if [tex]\( x \)[/tex] is false, then [tex]\( z \)[/tex] must be true.
- There is no implication rule that directly supports this statement from [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex].
D. [tex]\( z \Rightarrow x \)[/tex]:
- The statement suggests that if [tex]\( z \)[/tex] is true, then [tex]\( x \)[/tex] must also be true.
- There is no implication rule that supports this reversal of the initial implications.
By considering the transitive property of implication, the only statement that must be true given [tex]\( x \Rightarrow y \)[/tex] and [tex]\( y \Rightarrow z \)[/tex] is:
B. [tex]\( x \Rightarrow z \)[/tex].
