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

[moredumbimpossi]

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

Demonstrate this:

(p ∧ q) ↓ q ≡ ¬q

PLEASE.

 

[I like Serena]

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?

 

[moredumbimpossi]

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

 

[I like Serena]

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?