Commutation, Association and Distribution (Com, Assoc, Dist)

Three more rules of replacement let you reorder, regroup and spread out conjunctions and disjunctions. With them, together with De Morgan and Double Negation (DM, DN), you can reshape a line until an inference rule fits.

Commutation, association and distribution.
Rule	Form	Form
Com	$p \lor q$	$q \lor p$
Com	$p \land q$	$q \land p$
Assoc	$p \lor (q \lor r)$	$(p \lor q) \lor r$
Assoc	$p \land (q \land r)$	$(p \land q) \land r$
Dist	$p \land (q \lor r)$	$(p \land q) \lor (p \land r)$
Dist	$p \lor (q \land r)$	$(p \lor q) \land (p \lor r)$

Key idea

Com swaps the two sides of $\land$ or $\lor$. Assoc regroups three parts joined by the same connective. Dist spreads $\land$ over $\lor$, or $\lor$ over $\land$, like multiplication over addition.

Why it works

The order of the parts never matters for $\land$ and $\lor$, since "both" and "at least one" do not care which comes first. Grouping does not matter when all the connectives are the same: "$p$ and $q$ and $r$" means all three, however you bracket them. Dist is the logical version of $a(b + c) = ab + ac$: "$p$, and either $q$ or $r$" is true exactly when "$p$ and $q$" or "$p$ and $r$" is.

Distribution: the two columns match in all eight rows.

Worked examples

Worked example: Commutation

Apply Com to $(r \to s) \land p$.

The main connective is $\land$, so its two sides may swap.
The conditional inside is untouched.

Answer: $p \land (r \to s)$

Worked example: Assoc

Apply Assoc to $(p \land q) \land r$.

Three parts, all joined by $\land$.
Regroup to the right.

Answer: $p \land (q \land r)$

Worked example: Dist to free a disjunct

Premises $p \land (q \lor r)$ and $\neg (p \land q)$. Prove $p \land r$.

Line 3: $(p \land q) \lor (p \land r)$ by Dist 1.
Line 4: $p \land r$ by DS 3, 2.

Answer: Dist turns the premise into a disjunction that DS can use.

Common mistake

Using Com on a conditional (turning $p \to q$ into $q \to p$) or Assoc on mixed connectives (turning $p \land (q \lor r)$ into $(p \land q) \lor r$). Neither is allowed, and both change the meaning: the first is the converse, and the second is false when $p$ is false and $r$ is true. Com and Assoc work only with $\land$ and $\lor$, and Assoc only when all the connectives are the same.

These three rules rarely finish a proof on their own. Their job is to line formulas up. DS, for instance, needs the negation of one disjunct; if the disjunction has three parts, Assoc can group them so that the part you can deny stands alone on one side. Com does the same when a conclusion lists the parts in a different order from the line you have.

A tip for Dist: it is the rule to reach for when a line has $\land$ outside and $\lor$ inside (or the reverse) and you need the other arrangement. Read it in both directions: from $(p \lor q) \land (p \lor r)$ you can pull the common $p$ back out, giving $p \lor (q \land r)$.

Commutation, Association and Distribution (Com, Assoc, Dist)

Worked examples

Practice

Builds on

Leads to

Sources and further reading