Difficulty: Easy
Correct Answer: True
Explanation:
Introduction / Context:
This syllogism mixes class membership (Lamels are Signots) with an attribute (having buttons) and a color constraint (yellow Signots lack buttons). We test whether a universal exclusion follows.
Given Data / Assumptions:
Concept / Approach:
If every Lamel is a Signot that has buttons, and no yellow Signot has buttons, then a Lamel cannot be a yellow Signot because that would contradict “has buttons.” Therefore, Lamels must be non-yellow.
Step-by-Step Solution:
1) From S1: For any x, if x is Lamel ⇒ x is Signot and x has Buttons.2) From S2: For any x, if x is Yellow and x is Signot ⇒ x does not have Buttons.3) Suppose, for contradiction, some Lamel is Yellow: then it is a Yellow Signot with Buttons and without Buttons simultaneously, impossible.4) Hence, no Lamel is Yellow ⇒ statement is True.
Verification / Alternative check:
Venn-style reasoning: Lamels live entirely within the “Signots with buttons” region, while yellow Signots live entirely outside the “with buttons” region; the sets cannot overlap.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing “No yellow Signots have buttons” with “No Signots have buttons.” The color qualifier is essential; combined with S1, it still excludes yellow Lamels.
Final Answer:
True
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