#A knights and knaves puzzle that got me scratching my head

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wicked orbit
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The problem reads:

A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie.

You meet two inhabitants: Peggy and Zippy. Peggy tells you that “of Zippy and I, exactly one is a knight'. Zippy tells you that only a knave would say that Peggy is a knave.

Can you determine who is a knight and who is a knave?

Attached is a picture of my attempt at solving the problem. I ended up with the answer Peggy is a knight and Zippy is a knave. Is my answer correct? I used the truth table framework from this video from Mathematics Visual Proofs: https://www.youtube.com/watch?v=C6PeX4iKJbU&t=539s

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YouTube
Mathematical Visual Proofs
Knights, Knaves, and Propositional Logic [Discrete Math Class]

This video is not like my normal uploads. This is a supplemental video from one of my courses that I made in case students had to quarantine. I decided that it might be of general interest for people wanting to learn about propositional logic (mathematical propositions, logical connectives - "and", "or", and "not", truth tables, logical equivale...

▶ Play video
young horizonBOT
round flare
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Based on your definition for the statements: $\neg p$ and $\neg q$, do you know the meaning of $\neg q \rightarrow \neg p$?
What is the translation of that symbolic logic in English?

vast zodiacBOT
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The Mathematics Guy

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summer walrus
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What i feel :
let P = "Peggy is a knight"
let Z = "Zippy is a knight"

  1. Peggy's Statement :
    "Of Zippy and me, exactly one is a knight."
    Meaning : P ⊕ Z (exclusive‑or).

If  P is true (Peggy a knight), then Z must be false => (P, Z) = (T, F)

If  P is false (Peggy a knave), her utterance must be false, so P and Z have the same truth‑value. With P=F this forces Z=F.

Thus only two candidate worlds remain:

(T, F) — Peggy knight, Zippy knave.

(F, F) — both knaves.

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  1. Zippy's statement :
    "Only a knave would say that Peggy is a knave."

Formal Reading :
“For every person x, if x says 'Peggy is a knave', then x is a knave.

∀ x ¬ P → Knight ( x )

Such a universal statement is false precisely when there exists some truth‑teller who says “Peggy is a knave

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In case (T,F) Zippy is a knave who has just uttered a true statement—impossible for a knave.

Case (F,F) is perfectly consistent: both are knaves, Peggy's first remark is a lie, and Zippy's sweeping claim is also a lie because at least one truth‑teller can say “Peggy is a knave.”