MHB What is the simplified form of (p ∧ q) ↓ q using basic propositional logic?

AI Thread Summary
The discussion focuses on simplifying the expression (p ∧ q) ↓ q using propositional logic. Participants suggest using a truth table to analyze the expression, starting with different values for p and q. The key point is recognizing that the NOR operation can be expressed in terms of basic logical operations, specifically as ¬(p ∨ q). Ultimately, the goal is to demonstrate that (p ∧ q) ↓ q simplifies to ¬q. The conversation emphasizes understanding the underlying logic laws to achieve this simplification.
moredumbimpossi
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Please help me with this thing. I'm so frustrated I can't understand propositional logic

Demonstrate this:

(p ∧ q) ↓ q ≡ ¬q

PLEASE.
 
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moredumbimpossi said:
Please help me with this thing. I'm so frustrated I can't understand propositional logic

Demonstrate this:

(p ∧ q) ↓ q ≡ ¬q

PLEASE.

Hi moredumbimpossi, welcome to MHB!

What have you tried? Where are you stuck?

Simplest method to prove something like this, is to set up a truth table.
Let's start with p=0 and q=0.
What is (p ∧ q) ↓ q = (0 ∧ 0) ↓ 0 then?
 
I like Serena said:
Hi moredumbimpossi, welcome to MHB!

What have you tried? Where are you stuck?

Simplest method to prove something like this, is to set up a truth table.
Let's start with p=0 and q=0.
What is (p ∧ q) ↓ q = (0 ∧ 0) ↓ 0 then?

Hi., thanks for replying
This has to be reduced with the laws of logic
 
moredumbimpossi said:
Hi., thanks for replying
This has to be reduced with the laws of logic

Oh, okay.
Let's first get to basic operations then.
In general, we have $(a ↓ b) = \lnot (a \lor b)$ don't we? It's a NOR after all.
What do we get if we replace the NOR (↓) in (p ∧ q) ↓ q by those basic operations?
 
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