Question

If \(w=x\) and \(yx+z\) B) \(w yx-z\) D) \(w-x

Short answer

The correct proposition that holds in all cases is D) \( w - x < y - z \).

Step-by-step solution

Substitute w with x

The problem states \( w = x \). Substitute \( w \) with \( x \) in the proposed statements. This turns A) to D) into the following potential solutions: A) \( x + y > x + z \) B) \( x y < x z \) C) \( x - y > x - z \) D) \( x - x < y - z \)

Analyze each proposition

Now, check each of the propositions against the known fact that \( y < z \). A: \( x + y > x + z \). This equation simplifies to \( y > z \), which clashes with the given \( y < z \). So, A is false. B: \( x y < x z \). Assuming \( x \neq 0 \), this expression simplifies to \( y < z \), which matches with the given condition. So, B might be true. C: \( x - y > x - z \). This simplifies to \( y < z \), which matches the given condition. So, C might also be true. D: \( x - x < y - z \). This simplifies to \( 0 < y - z \) or \( z > y \), which matches the condition given. So, D might be true as well.

Check exception for Proposition B and Proposition C

Consider that in propositions B and C, it is implicitly assumed that \( x \neq 0 \). However, the exercise does not specify that. Thus, for \( x = 0 \), propositions B: \( 0 * y <0 * z \) (or \( 0<0 \)) and C: \( 0 - y > 0 - z \) (or \( 0 > 0 \)) are false.

Confirm the valid proposition

Since propositions A, B, C are proven to be false in at least one situation each, the only proposition that holds in all cases is D) \( w - x < y - z \). Hence, this is the correct option.