The relation ≤ and > on a boolean algebra are defined as: 𝑥≤𝑦 if and only if…
2018
The relation ≤ and > on a boolean algebra are defined as:
𝑥≤𝑦 if and only if 𝑥∨𝑦=𝑦
𝑥<𝑦 means 𝑥≤𝑦 but 𝑥≠𝑦
𝑥≥𝑦 means 𝑦≤𝑥 and
𝑥>𝑦 means 𝑦<𝑥
Considering the above definitions, which of the following is not true in the boolean algebra?
(i) If 𝑥≤𝑦 and 𝑦≤𝑧, then 𝑥≤𝑧
(ii) If 𝑥≤𝑦 and 𝑦≤𝑥, then 𝑥=𝑦
(iii) If 𝑥<𝑦 and 𝑦<𝑧, then 𝑥≤𝑦
(iv) If 𝑥<𝑦 and 𝑦<𝑧, then 𝑥<𝑦
Choose the correct answer from the code given below:
- A.
(i) and (ii) Only
- B.
(ii) and (iii) Only
- C.
(iii) Only
- D.
(iv) Only
Attempted by 105 students.
Show answer & explanation
Correct answer: D
Given: x<y ⇒ x≤y and x≠y.
Now examine each statement:
(i) If x≤y and y≤z, then x≤z.
Step 1: x≤y means x∨y = y.
Step 2: y≤z means y∨z = z.
Step 3: Then x∨z = (x∨y)∨z = y∨z = z, so x≤z.
✔ True (transitivity).
(ii) If x≤y and y≤x, then x=y.
Step 1: x≤y ⇒ x∨y = y.
Step 2: y≤x ⇒ y∨x = x.
Step 3: Since x∨y = y and y∨x = x, we have x = y.
✔ True (antisymmetry).
(iii) If x<y and y<z, then x≤y.
Step 1: x<y ⇒ x≤y by definition.
Step 2: Therefore, x≤y is true regardless of y<z.
✔ True.
(iv) If x<y and y<z, then x<y.
Step 1: x<y and y<z are given.
Step 2: The conclusion x<y is already assumed, not derived.
Step 3: The transitive property would give x<z, not x<y.
Step 4: Therefore, x<y does not follow as a logical consequence.
✘ Not true.
Thus, only (iv) is not true.
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