Equivalence relations, operator

Question

$\def\op{\mathbin{\#}}\def\R{\mathbin R}$Original question:

A logical operation between two propositions $p$ and $q$ is denoted as $p\op q$. It is only true when $p$ is true and $q$ is false; otherwise it is false. Let $R$ be a relation on the set of ordered pairs of propositions, where $R$ is defined as follows:

$\bigl((a,b),(c,d)\bigr) ∈ R$, if and only if $(a \land b) \op (c \vee d)$ is true.

Is R an equivalence relation?

For symmetric, the rule is $∀a,b ∈ B$ s.t. $a\R b$, $b\R a$ is true.

But when I do the relation, do I bring the operator (i.e. $\land$ and $\lor$) over also?

For example:

1) $(a \land b) \op (c \lor d) \to (c \lor d) \op (a \land b)$

2) $(a \land b) \op (c \lor d) → (c \land d) \op (a \lor b)$

which is the correct way of representing?

Answer 1

(a) A relation $R$ on the set $B$, is symmetric if and only if $\forall a,b \in B$, if $\;a\, R\, b$ is true, then $b\,R\,a$ is true. $\def\op{\mathbin{\#}}$

Here the elements of $B$ are ordered pairs of propositions.

$\bigl((a,b),(c,d)\bigr) \in R$, if and only if $(a \land b) \op (c \vee d)$ is true.

So you must show that $R$ is symmetric, you need to show that $(2)$, as you've written, holds. That is, you need to show that $$\forall (a, b) \in B,\; [(a \land b) \op (c \lor d)] \rightarrow [(c \land d) \op (a \lor b)].$$

Whether your relation is symmetric or not depends on the set on which the relation is defined. If it is defined on the set of all ordered pairs of all propositions, e.g., $(a, b)$, where $a, b\, \in \{\text{All possible propositions}\}$, then you should be able to easily find a pair of propositions that provides a counterexample to symmetry.

Note: that to determine whether $R$ is an equivalence relation on the set $B$ of pairs $(x, y)$, you need to also need to determine whether the relation is reflexive and transitive, as well as symmetric:

(b) The relation $R$ on the set $B$ is reflexive if and only if for all $(a,b) \in B$, $(a \land b) \op (a \lor b)$ is true.

(c) The relation $R$ is transitive if and only if for all $(a, b), (b, c), (c, d) \in B$, if $(a \land b) \op (b \lor c)$ and $(b \land c) \op (c \lor d)$ are both true, then $(a \land b) \op (c \lor d)$ is also true.

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Note that for $\op$, as defined, $(p \op q) \iff (p \land \lnot q) \iff \lnot(p\rightarrow q)$.

Answer 2

Way 2 is the correct way, of course.
Arguments of R are pairs of propositions, not the operations $\wedge$ and $\vee$ on that pairs.
So, R is not symmetric.