Intelligent Systems II · TEE · papers

Q1(c) Fuzzy set operations [5]

Re-examination · 31 Jan 2026

Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged)

Question. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y. [5 marks]

Answer

Given

The aligned X and Y memberships printed in this question.

To find

The four requested membership lists in the original element order.

Formula and why it applies

Use max for OR, min for AND, and 1−μ for NOT.

\mu_{X\cup Y}=\max(\mu_X,\mu_Y),\quad\mu_{X\cap Y}=\min(\mu_X,\mu_Y)

Complement both operands before applying the operator in ¬X∪¬Y or ¬X∩¬Y.

Solution

	c1	c2	c3	c4	c5
¬X	0.3	0.1	0.8	0.6	0.1
¬Y	0.4	0.7	0.2	0.3	0.9
X ∪ Y	0.7	0.9	0.8	0.7	0.9
X ∩ Y	0.6	0.3	0.2	0.4	0.1
¬X ∪ ¬Y	0.4	0.7	0.8	0.6	0.9
¬X ∩ ¬Y	0.3	0.1	0.2	0.3	0.1

Final answer. As in the last four rows. De Morgan check: ¬X ∪ ¬Y = 1 − (X ∩ Y) and ¬X ∩ ¬Y = 1 − (X ∪ Y) ✓.

Calculate every element separately

Element c1

\neg X:1-0.7=0.3\qquad\neg Y:1-0.6=0.4

X\cup Y:\max(0.7,0.6)=0.7

X\cap Y:\min(0.7,0.6)=0.6

\neg X\cup\neg Y:\max(0.3,0.4)=0.4

\neg X\cap\neg Y:\min(0.3,0.4)=0.3

Element c2

\neg X:1-0.9=0.1\qquad\neg Y:1-0.3=0.7

X\cup Y:\max(0.9,0.3)=0.9

X\cap Y:\min(0.9,0.3)=0.3

\neg X\cup\neg Y:\max(0.1,0.7)=0.7

\neg X\cap\neg Y:\min(0.1,0.7)=0.1

Element c3

\neg X:1-0.2=0.8\qquad\neg Y:1-0.8=0.2

X\cup Y:\max(0.2,0.8)=0.8

X\cap Y:\min(0.2,0.8)=0.2

\neg X\cup\neg Y:\max(0.8,0.2)=0.8

\neg X\cap\neg Y:\min(0.8,0.2)=0.2

Element c4

\neg X:1-0.4=0.6\qquad\neg Y:1-0.7=0.3

X\cup Y:\max(0.4,0.7)=0.7

X\cap Y:\min(0.4,0.7)=0.4

\neg X\cup\neg Y:\max(0.6,0.3)=0.6

\neg X\cap\neg Y:\min(0.6,0.3)=0.3

Element c5

\neg X:1-0.9=0.1\qquad\neg Y:1-0.1=0.9

X\cup Y:\max(0.9,0.1)=0.9

X\cap Y:\min(0.9,0.1)=0.1

\neg X\cup\neg Y:\max(0.1,0.9)=0.9

\neg X\cap\neg Y:\min(0.1,0.9)=0.1