Understanding absorption law

Question

I can't understand how absorption law is obtained. I get following steps.

$$a∨(a∧) = (a∧⊤)∨(a∧)$$ $$=(a∨a)∧(a∨b)∧(⊤∨a)∧(⊤∨b)$$ then,

I come up with $$=a∧(a∨b)∧⊤∧⊤$$ $$=a∧(a∨b)$$

But, I cannot get $\color{blue}{a∧(⊤∨)}$, as shown on Boolean Algebra: Axiom (1 2): ∧ distributes over ∨, therefore $\color{blue}a$.

Can you help me? I cannot obtain $\color{blue}{a∧(⊤∨)}$. Some people say in other answers in different questions, it is obtained by distribution law. However, what I got by this is the first equation.

Answer 1

Distribution, like any other boolean algebra law, is an equivalence principle, and so works both ways.

That is, given that $$P \land (Q \lor R) = (P \land Q) \lor (P \land R)$$

you can not only go from $$P \land (Q \lor R)$$ to $$ (P \land Q) \lor (P \land R)$$

but you can also go ‘back’, i.e. go from $$(P \land Q) \lor (P \land R)$$ to $$P \land (Q \lor R)$$

And yes, that doesn’t *feel * like ‘distribution’ … maybe more like ‘un-distribution’ or ‘factoring’ … but it is an application of Distribution nevertheless.

This is really not any different from basic high school algebra: we know that $3* (4 +7) = (3*4) + (3 * 7)$ … but that also means that $(3*4) + (3 *7) = 3 *(4 +7)$. Again, it can go both ways.

Answer 2

$$ a\land(\top\lor b) = (a\land \top)\lor (a\land b) $$ $$ = a \lor (a\land b) $$

Sometimes it helps starting with what you want to obtain

Answer 3

You can use a truth table to prove that $P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$ is a tautology.

Using natural deduction, you can use a simple proof by cases in each direction (32 lines).

In the forward direction, consider the cases $Q$ and $R$. In the backward direction, consider the cases $P\land Q$ and $P\land R$.

Answer 4

For the sake of variety, I offer proof the two formulas are logically equivalent via primitive inference rules...

First, proof that $p \vee (p \wedge q) \vdash p$

$\{1\} \:\:\:\:\: 1. p \vee (p \wedge q) \:\:\:\:\:$ premise

$\{2\} \:\:\:\:\: 2. p \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ Assumption

$\{3\} \:\:\:\:\: 3. p \wedge q \:\:\:\:\:\:\:\:\:\:\:\:\:\:$ Assumption

$\{3\} \:\:\:\:\: 4. p \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $3$ &E

$\{1\} \:\:\:\:\: 4. p \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $1,2,2,3,4$ vE

Second, proof that $p \vdash p \vee (p \wedge q)$

$\{1\} \:\:\:\:\: 1. p \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ premise

$\{1\} \:\:\:\:\: 2. p \vee (p \wedge q) \:\:\:\:\:$ $1$ vI

Thus, we have $p \vee (p \wedge q) \dashv \vdash p$