Question about Absorption Laws in Boolean Algebra

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The absorption laws in Boolean algebra state that p ∨ (p ∧ q) = p and p ∧ (p ∨ q) = p. A discussion arose regarding a partial absorption example, ¬q ∧ (¬p ∨ q) = ¬q ∧ ¬p, which is not part of the standard absorption laws. The conversation highlighted the connection between Boolean algebra and set theory, illustrating the laws using Venn diagrams. It was noted that the property p ∨ (¬p ∧ q) = p ∨ q is derived from the distribution of logical operators and is not considered fundamental. The discussion emphasized the importance of understanding these concepts through visual aids like Venn diagrams.
polyglot
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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?
 
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polyglot said:
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 said:
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 :
$$A \cup C = A, A \cap D = A$$
 
PeroK said:
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
 
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polyglot said:
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.
 
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