Question about Absorption Laws in Boolean Algebra

[polyglot]

Homework Statement

Absorption laws in Bollean algebra

Relevant Equations

¬q ∧ (¬p∨q)

According to my notes, the absorption law states that p ∨ (p ∧ q) = p, p ∧ (p ∨ q) = p
I have found a video where they were discussing a partial absorption such as ¬q ∧ (¬p∨q) = ¬q ∧ ¬p
This is not in my notes, but is this correct? specifically, is the terminology used to decribe this property the correct one?

 

[PeroK]

Homework Statement: Absorption laws in Bollean algebra
Relevant Equations: ¬q ∧ (¬p∨q)

According to my notes, the absorption law states that p ∨ (p ∧ q) = p, p ∧ (p ∨ q) = p
I have found a video where they were discussing a partial absorption such as ¬q ∧ (¬p∨q) = ¬q ∧ ¬p
This is not in my notes, but is this correct? specifically, is the terminology used to decribe this property the correct one?

I've never studied boolean algebra, so I look at these in terms of set theory. The absorption laws are simply:
$$A \cup (A \cap B) = A, \ A \cap (A \cup B) = A$$Which are obvious from a Venn diagram.

In the partial absorption you quote, note that only $\neg p$ appears, so you might as well replace that with $p$. In any case, straighteming out the logic and confusing order, this says:
$$A \cap (A' \cup B) = A \cap B$$Which is also obvious from a Venn diagram.

PS in this last one I took $\neg q \leftrightarrow A$ and $\neg p \leftrightarrow B$. $A'$ is the complement of $A$.

 

[PeroK]

I've never studied boolean algebra, so I look at these in terms of set theory. The absorption laws are simply:
$$A \cup (A \cap B) = A, \ A \cap (A \cup B) = A$$

More generally, if $C$ is any subset of $A$ and $D$ is any superset of $A$, then :
$$A \cup C = A, A \cap D = A$$

 

[polyglot]

I've never studied boolean algebra, so I look at these in terms of set theory. The absorption laws are simply:
$$A \cup (A \cap B) = A, \ A \cap (A \cup B) = A$$Which are obvious from a Venn diagram.

In the partial absorption you quote, note that only $\neg p$ appears, so you might as well replace that with $p$. In any case, straighteming out the logic and confusing order, this says:
$$A \cap (A' \cup B) = A \cap B$$Which is also obvious from a Venn diagram.

PS in this last one I took $\neg q \leftrightarrow A$ and $\neg p \leftrightarrow B$. $A'$ is the complement of $A$

Thanks - it is useful to look at it using a Venn diagram.
Is there a name for this property then p ∨ ( ¬p ∧ q) = p ∨ q

 

[PeroK]

Thanks - it is useful to look at it using a Venn diagram.
Is there a name for this property then p ∨ ( ¬p ∧ q) = p ∨ q

It's not fundamental, as it is a consequence of the distribution of $\vee$ over $\wedge$, complementation and identity laws.

 

[PeroK]

I'm using this as a reference, by the way:

https://en.wikipedia.org/wiki/Boolean_algebra