Question about Absorption Laws in Boolean Algebra

  • #1
polyglot
8
4
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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  • #2
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##.
 
  • #3
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 :
$$A \cup C = A, A \cap D = A$$
 
  • #4
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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  • #5
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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