Equivalence relation - help needed with proving

Question

let E be an Equivalence relation on A (non-empty group) is the next argument true or not, justify your answer:

$\exists x~(\forall y~(xEy)\lor \exists y~\forall z~(zEx\lor zEy)) \to \forall x~\exists y~\exists z((x\neq y)\land (x\neq z)\land (xEy\lor xEz))$

Let $A = \{1,2,3\}$ and $E = \operatorname{Id}a$

Is that a good counter-example?

To be clear, what is $\operatorname{Id}a$ meant to be? – Graham Kemp May 04 '18 at 10:48 — Graham Kemp, May 04 '18 at 10:48

score 0 · Answer 1 · answered May 04 '18 at 10:52

0

The formula is not valid: Let $A$ have only one element.

Your example doesn't satisfy the condition of the implication (has more than two equivalence classes), hence it's not a counterexample.

answered May 04 '18 at 10:52

Berci

ok, is the argument true? if so could you please show me how to prove it? – blabla May 04 '18 at 11:09
No, it's not true. A one element set provides a counterexample. – Berci May 04 '18 at 11:11

1 Answers1