How to deal with equivalences in proofs?

Question

There is a part I need clarification on regarding the use of equivalence and its symmetry.
From what I understand in regards to symmetry is that:
$ (p \equiv q) \equiv (q \equiv p) $. Given p and q are well formed formulas. Therefore: $((A\land B \equiv A \equiv B) \equiv (A \lor B)) \equiv ((A \lor B) \equiv (A\land B \equiv A \equiv B))$
Assuming p = $(A\land B \equiv A \equiv B)$ and q = $(A\lor B)$.
We know that:
$((A\land B \equiv A \equiv B) \equiv (A \lor B))$
is true since it is an axiom(golden rule). But does this mean
$((A \lor B) \equiv (A\land B \equiv A \equiv B))$
is a valid as a corollary since it is equivalent?
I know by common sense that the disjunction of A and B is not equivalent to its conjunction and likewise A or B but I don't know where the flaw is in the given pattern of thinking.

Answer 1

The so-called Golden Rule is "typical" of Equational Logic; thus I'll assume that your question must be "interpreted" in this context.

I'll refer to the textbook by George Tourlakis, Mathematical Logic (2008), which is devoted to the exposition of mathematical logic according to the "equational" style.

The axioms of the system are listed at page 42-43, and include the Golden Rule :

$A \land B \equiv A \equiv B \equiv A \lor B$.

Clearly, an axiom must be a tautology.

But we know that $A \land B, A \equiv B$ and $A \lor B$ are not equivalent, while

$(A \land B) \equiv (A \equiv B) \equiv (A \lor B)$

is a tautology.

So, where is the "trick" ?

What is the "utility" of the above "equation" if its "sub-formulae" are not equivalent ?

The "trick" is in the parentheses; the axiom as stated in Tourlakis' book is without parentheses. If we want to restore them, we must follow a rule.

We have [see page 15] the following rule regarding the priority (or precedence) between connectives :

Higher-priority connectives bind before lower-priority ones. That is [...] the higher-priority connective "glues" first.

The order of priorities (decreasing from left to right) is agreed to be :

$\lnot, \land, \lor, \rightarrow, \equiv$.

We must also note the comment in footnote 48 [page 42] : "Recall that brackets associate from right to left".

Now, if we want to restore the parentheses into the Golden Rule, we must reintroduce them consistently : $\land$ and $\lor$ have higher-priority.

Thus, they "glue" first, and the formula becomes :

$(A \land B) \equiv A \equiv B \equiv (A \lor B)$.

But all the occurrences of $\equiv$ have obviously the same priority; thus, we must insert parentheses from right to left, with the result :

$(A \land B) \equiv (A \equiv B \equiv (A \lor B))$.

As you can easily verify, this is a tautology.

When $(A \equiv B)$ has the same value of $(A \lor B)$ ? Only when both $A$ and $B$ are true; i.e. exactly when $(A \land B)$ hodls.

But it does not assert that $A \land B$ is equivalent to $A \lor B$, nor to $A \equiv B$.

Of course, $\equiv$ is symmetric; thus, the above formula is equivalent to : $(A \equiv B \equiv (A \lor B))\equiv (A \land B)$.

But we can also "permute" the subformulae; when $(A \land B)$ has the same value of $(A \lor B)$ ? When $A$ and $B$ are both true or both false; i.e. when $(A \equiv B)$ hodls. Thus, we have also :

$((A \lor B) \equiv (A \land B)) \equiv (A \equiv B)$.

We must note that in the context of equational logic a derivation is basically a chain of equivalences. The axioms must be used "in collaboration" with the rules of inference [see page 40] which include Equanimity [an "equational" form of detachement rule] :

from $A$ and $A \equiv B$, derive $B$.

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Note

The "equational" style looks different from the traditional exposition of mathematical logic, but it "works" basically in the same way.

See Herbert Enderton, A Mathematical Introduction to Logic (2nd - 2001), page 33 : the priority rules for connectives are the same and when one connective symbol is used repeatedly, the convention for grouping is to the right.

Answer 2

"We know that: ((A∧B≡A≡B)≡(A∨B)) is true since it is an axiom(golden rule). But does this mean ((A∨B)≡(A∧B≡A≡B)) is a valid as a corollary since it is equivalent?"

Yes. A quick way to verify this comes as to look at the truth table for ≡.

 ≡      False   True
False   True    False
True    False   True

Now suppose that (p≡q) is true and "p" is true also. p as true corresponds to the second row of the truth table. (p≡q) as true corresponds to the (false, false) position and the (true, true) position of the truth table. So, (p≡q) as true and "p" as true overlap in the lower right hand quadrant of the truth table. In this case, q is true also. Consequently, it follows that if (p≡q) is true, and if "p" is true also, then "q" is also true.

Now, you know that [(a≡b)≡(b≡a)] is a tautology. It follows that if you have any axiom or theorem of the form (p≡q) which is a tautology, (q≡p) can get deduced also.