Can I get some help on this one? Any solutions?
Let $≺$ be a relation on a set $P$ that is reflexive and transitive. Define the relation $∼$ on $P$ by
$p ∼ q$ iff $p ≺ q ∧ q ≺ p$.
(a) Show $∼$ is an equivalence relation.
(b) Given any $p ∈ P$, let $[p]$ denote the $∼$-equivalence class of $p$. Define $(P/∼, ⪯)$ to comprise the set $P/∼$ of $∼$-equivalence classes on which $x ⪯ y$ iff $∃p, q. (x = [p]) ∧ (y = [q]) ∧ p ≺ q$.
Show $(P/∼, ⪯)$ is a partial order.
Ok, this is what I got:
An equivalence relation is:
Reflexive:$p~p → p≺p ∧ q≺q$
Which is true since the relation $≺$ is reflexive.
Symmetric:if $p~q$ then $q~p → p≺q ∧ q≺p$ then $q≺p ∧ p≺q$ Which must be true since the logical operator “and” do not stipulate that the statements are fixed to either side of the operator.
Transitive:if $p~q$ and $q~u$ then $p~u → p≺q ∧ q≺p$ and $q≺u ∧ u≺q$ then $p≺u ∧ u≺p$ Which is true since the relation $≺$ is transitive.
Is this on the right track. Have no clue what to do in part b).
[p] <= [p] because p <= p
If x <= y, y <= x, then exists p,q,r,s with
x = [p], y = [q], y = [r], x = [s], p <= q, r <= s, p ~ s, q ~ r.
Show p <= r, r <= p; x = [p] = [r] = y.
If x <= y, y <= z, then exists p,q,r,s with
x = [p], y = [q], y = [r], z = [s], p <= q, r <= s, q ~ r.
Show p <= s; x = [p] <= [s] = z.