# Q1(c) Fuzzy union, intersection and complements [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-reexam/q1c

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-reexam--q1c/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q1(c)** [5 marks · 2024-06-25]

A={(x₁,.7),(x₂,.3),(x₃,.9),(x₄,.1)} and B={(x₁,.2),(x₂,.5),(x₃,.7),(x₄,.4)}. Find (1) A∪B, (2) A∩B, (3) Ā∩B̄, (4) Ā∪B̄.

### Answer

#### Given

Memberships A and B as printed; use the listed element order.

#### To find

Union, intersection and the requested operations on the complements.

#### Formula and why it applies

Membership is a degree between 0 and 1. Union means choose the larger membership; intersection means choose the smaller. Complement means subtract the membership from 1.



$$
\mu_{A\cup B}(x)=\max(\mu_A(x),\mu_B(x)),\quad \mu_{A\cap B}(x)=\min(\mu_A(x),\mu_B(x)),\quad \mu_{\bar A}(x)=1-\mu_A(x)
$$



#### Solution

##### Step 1. Compute complements before combining them

At x₁: 1−0.7=0.3, and 1−0.2=0.8. Never subtract the element label; subtract its membership.

##### Step 2. Work element by element

| Element | A | B | 1−A | 1−B | max (A, B) | min (A, B) | max(1−A,1−B) | min(1−A,1−B) |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| x₁ | 0.7 | 0.2 | 0.3 | 0.8 | 0.7 | 0.2 | 0.8 | 0.3 |
| x₂ | 0.3 | 0.5 | 0.7 | 0.5 | 0.5 | 0.3 | 0.7 | 0.5 |
| x₃ | 0.9 | 0.7 | 0.1 | 0.3 | 0.9 | 0.7 | 0.3 | 0.1 |
| x₄ | 0.1 | 0.4 | 0.9 | 0.6 | 0.4 | 0.1 | 0.9 | 0.6 |

##### Step 3. Read the requested columns in the original order

Keep the labels unchanged. Only the membership next to each label changes. De Morgan checks: complement of the union equals intersection of complements; complement of the intersection equals union of complements.

**Final answer.** (1) A∪B=[0.7, 0.5, 0.9, 0.4]; (2) A∩B=[0.2, 0.3, 0.7, 0.1]; (3) Ā∩B̄=[0.3, 0.5, 0.1, 0.6]; (4) Ā∪B̄=[0.8, 0.7, 0.3, 0.9].