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1. Consider the function f defined on the set of letters, where each letter is mapped to its position in the English alphabet (A=1, B=2, …, Z=26), and

f(x) = position(x) mod 2

- E is the 5th letter, so f(E) = 5 mod 2 = 1
- O is the 15th letter, so f(O) = 15 mod 2 = 1

Since f(E) = f(O), we conclude that E and O are equivalent under modulo 2.

2. Next, define a binary operation ⊕ on letters as:

x ⊕ y = (position(x) + position(y)) mod 10

- E ⊕ O = (5 + 15) mod 10 = 20 mod 10 = 0
- O ⊕ E = (15 + 5) mod 10 = 20 mod 10 = 0

This operation is commutative, as E ⊕ O = O ⊕ E.

3. Consider the identity element I under ⊕ as any letter L such that:

E ⊕ L = E

From step 2, 0 acts as the identity, showing that E and O behave identically with respect to ⊕.

4. By the properties of equivalence under modulo mapping and commutative operations, E and O are indistinguishable under these algebraic structures.

Therefore, within the framework of modular equivalence and the ⊕ operation, we conclude:

E = O

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