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Professor Citachka
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Unit quiz: Connectives and conditional inference

Check understanding, separately from reading

5 questions · 80% to pass · no timer. This quiz checks the lessons in this unit. These authored questions assess recognition and application of the taught distinctions, not professional qualification. You may review the lessons and retry. Repeat attempts reuse the question bank; a remembered answer is not proof of transfer to a new situation.

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Question 1

Let P mean “the archive is open” and Q mean “the manuscript is accessible.” The argument “If P then Q; P; therefore Q” relates two complete claims. Replacing P with “archive” would lose the assertion that can be true or false. Before evaluating the argument, write the key for both letters.

What should the letter P represent in this translation?

Question 2

A reading requirement says “submit a commentary or a translation.” Under inclusive or, submitting both satisfies it. Under exactly-one instructions, both violates it. The cases P true/Q false and P false/Q true satisfy either rule; P true/Q true is the case that distinguishes them.

Which case separates inclusive or from exactly one?

Question 3

Formalize “If the seal is broken, the indicator is red.” A broken seal with a non-red indicator refutes that conditional. An intact seal with a red indicator does not: another mechanism might make it red. The formula alone says nothing about that mechanism.

Which observation falsifies the material conditional?

Question 4

If a file is encrypted, it is unreadable without the key. This file is unreadable. Concluding that it is encrypted overlooks corruption. A corrupted, unencrypted file makes both premises true and the conclusion false. The problem is the inference, even if this particular file later turns out to be encrypted.

What does the corrupted-file countermodel establish?

Question 5

If a shape is a square, it has four sides. A triangle does not have four sides, so it is not a square: modus tollens. A non-square rectangle still has four sides, showing why “not square, therefore not four-sided” fails.

Which inference is licensed by the stated square conditional?

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